text:
Print the phrase "And she said, 'But that's his.'" using only the alphabet
Print the phrase And she said, 'But that's his.' using only the following characters: abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ  No punctuation or non-alphabetic characters whatsoever.  You can use whatever programming language you want.  Whitespace is completely allowed.  Shortest program wins.

What about whitespace in output? (leading/trailing?)

Output must be exactly what was specified. Usually an optional trailing newline is allowed.

Darn, my esolang can't complete because there's no ability to produce output with only a-zA-Z. In theory I could use write and Eval to create the necessary instructions, but none of +-*,%'" can be constructed without using (at least) one of +-*,%'"0-9.

What about control characters? (escape, control, alt, etc.)?

(programmer-of (language 'lisp)) dislikes this.

Have to admit, I didn't think this was particularly interesting at first, but the combination of repeated and unique characters really made it something fun to optimize (especially on a stack language!). Very nice.

Can you clarify if extra whitespace in the output is allowed?  Like extra trailing newlines?  Or only whitespace in the source, plus alphabetic characters.  There's a Befunge answer that prints with extra an trailing newline.

Whitespace, 417 414 349 265 bytes
265 bytes thanks to Kevin Cruijssen

Try it online!
Explained:
[S S T  T   T   T   T   T   T   N
_Push_-63_'][S S T  T   T   T   S S S N
_Push_-53_.][S S S T    T   S T N
_Push_13_s][S S S T T   N
_Push_3_i][S S S T  S N
_Push_2_h][S S T    T   S S S T T   S N
_Push_-70_space][S T    S S T   T   N
_Copy_0-based_3rd_s][S T    S S T   T   S N
_Copy_0-based_6th_'][S S S T    T   T   S N
_Push_14_t][S S T   T   S T N
_Push_-5_a][S S S T S N
_Push_2_h][S T  S S T   S N
_Copy_0-based_2nd_t][S T    S S T   T   S N
_Copy_0-based_6th_space][S T    S S T   N
_Copy_0-based_1st_t][S S S T    T   T   T   N
_Push-15_u][S S T   T   S S T   S S N
_Push_-36_B][S S T  T   T   T   T   T   T   N
_Push_-63_'][S T    S S T   S S N
_Copy_0-based_4th_space][S S T  T   T   T   S T S N
_Push_-58_,][S S T  T   S N
_Push_-2_d][S S S T T   N
_Push_3_i][S S T    T   S T N
_Push_-5_a][S S S T T   S T N
_Push-13_s][S T S S T   S T N
_Copy_0-based_3rd_space][S S T  T   N
_Push_-1_e][S S S T S N
_Push_2_h][S T  S S T   T   N
_Copy_0-based_3rd_s][S T    S S T   T   N
_Copy_0-based_3rd_space][S S T  T   S N
_Push_-2_d][S S S T S S S N
_Push_8_n][S S T    T   S S T   S T N
_Push_-37_A][N
S S N
_Create_Label_LOOP][S S S T T   S S T   T   S N
_Push_102][T    S S S _Add][T   N
S S _Print_as_character][N
S N
N
_Jump_to_Label_LOOP]

Whitespace is completely allowed. I see you have taken this literally.

You beat me to it.. Quite a few things can be golfed, though. :) You can remove the trailing NNN to exit, since it already stops with an error when you're doing the add before the print_char, so it won't even come after the Jump_to_Label. Also, why store the 63 at the start and retrieve it in the loop? You could just push it before adding instead. And, why is the Label-nr TTSSSSTN? A Label can even be empty, so just NSSN to create the label and NSNN to jump to the label is sufficient when you're only using one Label.

318 bytes with the changes I proposed above. Here is the same program with added highlighting. And how did you deducted the constant value 63? I'm not 100% sure it's the shortest constant possible here. If it is, something is wrong with my constant generating program I wrote for an earlier challenge. :)

Yep, I was right. Constant 102 is the most efficient: 281 bytes (or here with highlighting). (NOTE: I've also used one copy to save 4 bytes for the space between ehs dnA (copied from the space between dias ehs).

Ok, now I'm done. :) 265 bytes (or here with highlighting). Added some additional copies. (Here the relevant Whitespace tip.)

Perl 5, 133 102 95 bytes
s qqAnd she saidZ ZBut thatZs hisZZGjGGfq x
s qZqchr oct oct oct ord chopqge x
y qGjfqqdx
print

Try it online!
Explanation:
Regexes, print, and chop all apply to the variable $_ by default.
s qqAnd she saidZ ZBut thatZs hisZZGjGGfq

Replaces the empty string with And she saidZ ZBut thatZs hisZZGjGGf.
s qZqchr oct oct oct ord chopqge

Replaces each Z with the result of evaling chr oct oct oct ord chop. This removes the last character of $_, takes its keycode, interprets it as octal thrice, and converts it back to a character. For example, j → 106 → 70 → 56 → 46 → ..
Due to the way replacement works, the modifications to $_ that happen while evaluating the replacement are lost, so $_ is now And she said, 'But that's his.'GjGGf.
y qGjfqqd

Deletes all G, j, and f in $_.

Not knowing Perl, it looks like you just tried to type the sentence, but got in multiple fights with your cat in the process

This may be the most beautiful piece of code I've ever seen written on this website, and I say that as someone who knows Perl.

Related, but you can replace print with say for -2 characters. Current meta consensus says command line flags like -M5.010 don't count toward byte count.

><>, 916 915 903 bytes
At first I thought a solution in ><> was impossible, but then I realized... who needs conditionals or logic control? :D
fffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffloffffffffffffffffffffffffffffffffffffffffffffflopppgloppppppppppppppppppppppggloffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffflopppggloploppppppppppppppppppppppploffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffflopppppplofffffffflopggloppppppppppppppppppgglopppplofffffffloffffffffffffffffffffffffffflofffffffffffffffffffffffffffffffffffffffffffffffffffloglopppppppppppppppppppppppppppplofffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffflopppploppgloffffffffffffffffffflopppppppppppppppppppppppppgglofffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffflopppppppppppppppppppppppppppgglofffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffflofloffffffffffloppppppppppppppppppppppploppgloio

Try it Online
I repeatedly push numbers (the number 15) to the stack, then push the length of the stack and print the character with that ASCII value. If I need to shrink the length of the stack, I shrink the stack three values at a time using p, or one at a time using g if I'm within three of the target. The program ends by calling i (input), which pushes a -1 since there's no input, then prints it to cause an error.
This is the Python 3 program I used to create the solution once I'd thought of how to do it:
s = "And she said, 'But that's his.'"
L = [0]+[ord(c) for c in s]
#print(L)
M = L[1:]+[0]
D = [x[1]-x[0] for x in zip(L,M)]
#print(D)
while D:
	n=D.pop(0)
	if not D:print('io',end='');exit()
	if n>0:print('f'*n,end='lo')
	else:
		while n<-2:print('p',end='');n+=3
		print('g'*-n,end='lo')

Try it Online

This is the proper counterpoint to the whitespace solution! Have a +1 from me, I also appreciate the generating program...

The last section can be removed entirely with careful use of p: create a 59 on the stack while moving up from the space to the s in said, then place it in the code on the way down from d to ,.  (Note that (15,15) has 15 by this point.)

Actually, you can abuse l and p to put several useful characters in the (10,10)-(15,15) rectangle, then retrieve them with g where its more convenient.

That's a really good idea. It's so hard to think of ><> in this way. It's like coding with BF

I posted my own answer: https://codegolf.stackexchange.com/a/186916/69059

Cool answer! I did something similar to this in a challenge to print a string with the fewest distinct characters possible: https://codegolf.stackexchange.com/a/219630/100146

8086 Assembly on IBM PC, 1463 845 664 bytes
Clarification: The actual assembly language source is the entry, not the generated machine code.
The difficulty is that most x86 instructions (like ADD, SUB, JMP, conditional jumps, memory access) have two arguments and thus need a comma, or need a memory address. So, we can't use addition, subtraction, ifs, or loops!
During my first attempt I was able to "construct" numbers using a combination of increment, decrement, multiplication, division, byte-tricks, and the obscure BCD instructions (like AAA, DAS). After that, I realised that this idea could be used to create self-inspecting and self-modifying code.

Attempt 1. (1463 bytes)
Used the available instructions to construct ASCII codes and the 0xb800 address of the screen buffer. The calculation of each character in the sequence was golfed by hand.
Attempt 2. (not complete)
Realised that there is an opcode for every integer in the range 0x40-0x5f. This range includes A-Z. So for example INC CX corresponds to 0x41 = 'A'. (This opcode table is very handy.)
I attempted to construct 3 "data" strings, and layer them on top of one another. The first as-is (uppercase), the second "shifted" into the 0x60-0x7f zone (lowercase) and the last "shifted" into the 0x20-0x3f zone (punctuation).
Self-modifying code would produce a loop or three to iterate over the data.
Attempt 3. (845 bytes)
Like previous approach but to cut down on data, the string would only be encoded once, with "control characters" mixed in to switch character sets.
Attempt 4. (664 bytes)
How to get rid of the control characters which require a whole lot of patched instructions to deal with branching? Given that only two uppercase letters are used, I wondered whether I could "flip" the opcode table to encode lowercase letters using the 0x40-0x4f range and punctuation using the 0x90-0x9f range (by subtracting from 0xc0). The "A" and "B" could get put in separately.
However only half the opcodes in the 0x90-0x9f range are usable and they didn't line up with the needed ones. Then I thought maybe I could shuffle them around using a XOR, and I found one that worked. And here it is.

Golfed:
REP LODSB
PUSH CX
PUSH CX
POP AX
INC CH
PUSH CX
POP DI
DEC AX
DEC AX
REPNE SCASB
REPNE SCASB
PUSH DI
REPNE SCASB
PUSH DI
REPNE SCASB
PUSH DI
POP SI
POP DI
DEC DI
LODSB
NOT AL
STOSB
POP CX
DEC CH
LODSB
NOT AL
STOSB
LODSB
AAA
STOSB
INC DI
LODSB
NEG AL
STOSB
LODSB
NOT AL
PUSH AX
PUSH AX
INC SP
POP ES
INC SP
POP DI
LODSB
NOT AL
PUSH AX
POP BX
NEG AL
STOSB
INC DI
LODSB
DEC AL
NEG AL
DIV BH
PUSH AX
POP DI
LODSB
STOSB
RET
DEC BL
PUSH CS
STOSB
PUSH DS
INC DI
INC AX
POP SI
PUSH SP
NOP
INC BP
POP AX
PUSH DI
NOP
INC BP
PUSH BX
POP BX
PUSH SP
PUSHF
NOP
CWD
PUSH DX
INC DI
INC SP
NOP
INC SP
POP AX
PUSH BX
INC SP
CWD
INC BP
NOP
POP AX
POP BX
INC BP
SAHF
CWD
SCASB
INC DX

Assemble with
nasm golf.asm -o golf.com

and run in DOSBOX (run CLS first). Looks like this:

Commented:
; ASSUME DS = ES = CS
; ASSUME IP = 0x0100
; true for .COM file

; We treat 0xFE as a special marker that we scan for
; This marks our patch zone and the start of our data

; We also use it as a cheap trick to get a constant 0x1f
; into CX

; 0xFE is the first byte of INC or DEC instructions
; that operate on half-word registers (AL, BL, CH etc.)
; WATCH OUT! Adding these breaks the scan

; Can't assume any register contains zero
; so use this trick to zero out CX
REP LODSB

PUSH CX ; needed later

; zero AX
PUSH CX
POP AX

INC CH
PUSH CX
POP DI ; 0x100, where our code starts

DEC AX
DEC AX ; AL = 0xFE, our marker (AH = 0xFF)

REPNE SCASB ; skip the INC CH above
REPNE SCASB ; find the DEC CH located at 0x10E

; we will later need 0xF, the char count minus the 'A'
PUSH DI ; DI = 0x10F

REPNE SCASB ; find the patch position
PUSH DI

REPNE SCASB ; find the next 0xfe; our data section
PUSH DI
POP SI ; load data from here

POP DI ; store data to the patch position
DEC DI

; patch in XOR
; XOR is 0x34, start with 0xCB
; second byte of DEC BL is 0xCB
LODSB
NOT AL
STOSB

POP CX ; get 0x0f in CX for our strlen
DEC CH

; patch in our XOR arg
; it is 0xF1 (take 0x0E and NOT it)
LODSB ; 0x0E (PUSH CS)
NOT AL
STOSB

; ADD is 0x00 (take 0xAA, apply weird AAA behaviour)
; this also zeroes AH
LODSB ; 0xAA (STOSB)
AAA
STOSB

INC DI ; skip next instruction byte

; LOOP is 0xE2
LODSB ; 0x1E PUSH DS
NEG AL
STOSB

; get b800 in ES (address of screen buffer)
; first get 0x47 in AL (INC DI)
LODSB  ; get 0x47 (INC DI)
NOT AL ; NOT 0x47 = 0xb8
; AX = 0x00b8 (bytes backwards)

PUSH AX
PUSH AX
; stack contains 0xb8 0x00 0xb8 0x00
; stack off-by-1 trick
INC SP
; now POP gives you 0xb800
POP ES
INC SP ;and clean up after ourselves

; store 0 in DI ***** PUSHED AT START OF PROGRAM ***
POP DI

LODSB ; get our magic 0xC0 (0x40 INC AX)
NOT AL
PUSH AX
POP BX

NEG AL ; NOT and +1 to get 0x41 ("A")

; charloop:
STOSB
INC DI
LODSB
DEC AL ; XOR
NEG AL ; modify this into an ADD AL, BL
DIV BH ; modify this to LOOP back to charloop

; doesn't print the last character
; but the last character turns into the address where 'B'
; is supposed to go

PUSH AX
POP DI
LODSB ; "B"
STOSB

; graceful exit this time ;)
RET

; *** DATA SECTION ***

         ; PURPOSE

DEC BL   ; 0xFE marks data section, 0xCB for XOR
PUSH CS  ; for XOR arg
STOSB    ; for ADD
PUSH DS  ; for LOOP
INC DI   ; 0x47 -> for 0xb800

INC AX   ; for magic number but also "A"

POP     SI ;n
PUSH    SP ;d
NOP        ;
INC     BP ;s
POP     AX ;h
PUSH    DI ;e
NOP        ;
INC     BP ;s
PUSH    BX ;a
POP     BX ;i
PUSH    SP ;d
PUSHF      ;,
NOP        ;
CWD        ;'
PUSH    DX ;B
INC     DI ;u
INC     SP ;t
NOP        ;
INC     SP ;t
POP     AX ;h
PUSH    BX ;a
INC     SP ;t
CWD        ;'
INC     BP ;s
NOP        ;
POP     AX ;h
POP     BX ;i
INC     BP ;s
SAHF       ;.
CWD        ;'

SCASB     ; treated as char but turns into screen address!
INC DX    ; "B"

Hm. I am getting different .COM files from the two assembly sources, starting at offset 0x3e. Edit - Nvm found the difference: Line 117 in the commented version is INC AX while the uncommented is INC AL.

I want to see a fully alphabetic binary instead. :-)

I was initially thinking the answer was the resulting machine code, but it actually is the asm source. We can't use add/sub because 2-operand instructions need a comma, not because of their opcodes.  I thought the diff @gastropner saw was a bug because the 1-byte short form of inc ax is 0x40 (ASCII @), while the 2-byte (opcode + modrm) inc al is FE C0. but neither are ASCII.  inc any other 16-bit register is fine: the 1-byte short forms of inc/dec use the 0x40 .. 0x4f range of opcodes (with register number in the low 3 bits).  AX = 0, so we get 0x40.  ASCII A is 0x41.

(I normally golf in x86 machine code, not asm source.  IDK if it would be feasible to write a machine-code answer.  It might need to be self-modifying code to create the opcode for syscall (x86-64) or int 21h (DOS printing).  Oh, anatoly has already posted such an answer :)

Shouldn't the code end with a RET or otherwise exit cleanly? Or does NASM handle this for you?

@Oliphaunt the challenge didn't say anything about terminating... but yes you could add a RET.

If you are comfortable locking down NASM as the assembler of choice, you can create labels by doing label1 db on a line of its own. It will yield a warning, but no error.

@gastropner well that would just make it too easy. :P Didn't know about that, thanks! Maybe I should rename my language as "something you can feed into DEBUG.COM". Which incidentally I used to debug this. xD

@PeterCordes now this one is self-modifying!

@Artelius: cool.  And re: language: every assembler is basically a different language from a family of related languages, if you're talking about the asm source.  So the language of this answer is  8086 IBM PC assembly language (DEBUG.COM) IMO.  "Porting" it to other assembly dialects may be trivial or not :P  Or after your last update, "8086 IBM PC assembly language (NASM)".

This would be a better answer if you explain how it actually prints the string on the screen.  I see a b800 in there, the segment value for the base of VGA memory, assuming text mode.  A quick mention of that in the English text of how the answer works would be a good idea.

A better table for opcodes is referenced here: https://stackoverflow.com/a/32487675/509868

@anatolyg I beg to differ. Doesn't reference the actual registers or the argument format.

Perl 6, 1299 1272 1220 1215 bytes
Thanks to Grimy for -27 bytes.
-52 bytes because we didn't need the rabbit ears in the first place.
Thanks to Jo King for -5 bytes.
print chr flip chars i x chars i xx pi
and print lc chr chars NaN x chars cis pi
and print lc chr chars e x e x e
and print chr chars i x e x e x e
and print lc chr flip chars exp i
and print lc chr chars NaN x tau x e x e
and print chr chars chop NaN x e lcm chars e
and print chr chars i x e x e x e
and print lc chr flip chars exp i
and print lc chr flip chars i x chars i xx pi
and print chr chars False x pi x ceiling tau
and print lc chr chars e x e x e
and print chr chars i xx chars NaN x pi
and print chr chars i x e x e x e
and print chr chars chop False x e x e x e
and print chr chars chop NaN xx chars e
and print lc chr chars e x chars False
and print lc chr chars chop e x chars False
and print chr chars i x e x e x e
and print lc chr chars chop e x chars False
and print lc chr chars NaN x tau x e x e
and print lc chr flip chars i x chars i xx pi
and print lc chr chars chop e x chars False
and print chr chars chop False x e x e x e
and print lc chr flip chars exp i
and print chr chars i x e x e x e
and print lc chr chars NaN x tau x e x e
and print chr chars False x pi x ceiling tau
and print lc chr flip chars exp i
and print chr chars NaN xx tau x e
and say chr chars chop False x e x e x e

Try it online!
Outputs the string with a trailing newline. If you don't want that, then replace the last say with a print. You can also replace the newlines in the source with spaces.
Explanation
This code prints the string character by character. Each character is formed by feeding the appropriate character code into the chr function and lowercasing it with lc if needed.
Currently, all of the values are generated by generating a string with the correct number of characters in it; in some cases, the number of characters is the reverse of the target character code. It should be theoretically possible to use mathematical functions such as log and exp directly, but I didn't find it very easy to use those.
For use as numbers, we have e, pi and tau; in the right side of x or xx, they are implicitly floored. They all have 17 characters in their string representations, so we use e for minimal character count. We also have i (4 characters), False (5 characters) and NaN (3 characters). We can multiply string lengths with x; xx multiplies one plus the string length by the right hand side and adds one. chop removes one character from the string in case we're one away from the target.
The print statements are chained together using and, which has quite a low precedence. It's almost a miracle it exists; otherwise, we would have to use illegal semicolons.
I found the expressions for the characters by hand. It might be worth searching for them programmatically to find shorter expressions.

1078 bytes

By the way, @JoKing, did you search for the shorter expressions by hand or did you use a program to help?

By hand I'm afraid. An algorithmic way probably wouldn't be too hard

242 bytes

@Grimy  Clever, but unfortunately only possible because the Zero Width Space you're using is not a whitespace character

Width, 66 64 bytes
QaaGmwmiimaGcwWiimawAGawmfciiiGaFmAmFiimFGcwAmFiGmaiiGcGamafFiGQ

Try it online!
Prints to debug. To print to stdout, append ww to the end of the code, which pops and outputs the top of the stack.
Explanation
In Width, each letter correlates to a number, based on how "wide" it is, according to this table. This assigns each letter a number from 0 to 9. Then, those numbers are used to actually execute code.
In particular, a letter than matches 7 will start a string literal. It will read sets of two letters at once, until it reads the original letter again. Each set of two letters will be converted to their width numbers, read as a decimal number between 0 and 99, and the character they equal will be their index in the following string:
 !"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~\n\t

For example, the index of ! is 1, so 01 will be the right width numbers. Thus, if, iI, jt, etc will all correlate to a string literal of !.
In this case, I translated the 31 characters of the required output to appropriate letters, using Q as the quotation marks. The top of the stack is printed to debug when the program finishes.

This is the shortest one yet.  I think you could win!

Too bad how this language is presented in a monospaced font…

x86 machine code (32-bit), 256 bytes
When I print my code on my codepage 437 console, I see the following:
j XI a I a I a jbZ      Q fiQ Gf a f    Q I a I a I a I a h hisZ        Q I a I a I a I a hBP  Z        Q iQ
y       Q  a I a I a I a h thaZ Q I a I a I a Ih ButZ   Q  a I a I a I a fhu fZf        Q iQ g  S       Q  a I a I a I a hsaidZ Q I a I a I a I a hshe Z        Q I a I a I a I a hAnd Z        Q TZBX b

This contains some whitespace characters, so here is the same code when I replace all tab characters by → and all nonbreaking space characters (with code 255) by *:
j XI a I a I a jbZ→Q fiQ Gf a f→Q I a I a I a I a h hisZ→Q I a I a I a I a hBP  Z→Q iQ →→y →Q  a I a I a I a h thaZ→Q I a I a I a Ih ButZ→Q  a I a I a I a fhu fZf→Q iQ g→S →Q  a I a I a I a hsaidZ→Q I a I a I a I a hshe Z→Q I a I a I a I a hAnd Z→Q TZBX*b*

Hexdump:
6a 20 58 49 20 61 20 49 20 61 20 49 20 61 20 6a
62 5a 09 51 20 66 69 51 20 47 66 20 61 20 66 09
51 20 49 20 61 20 49 20 61 20 49 20 61 20 49 20
61 20 68 20 68 69 73 5a 09 51 20 49 20 61 20 49
20 61 20 49 20 61 20 49 20 61 20 68 42 50 20 20
5a 09 51 20 69 51 20 09 09 79 20 09 51 20 20 61
20 49 20 61 20 49 20 61 20 49 20 61 20 68 20 74
68 61 5a 09 51 20 49 20 61 20 49 20 61 20 49 20
61 20 49 68 20 42 75 74 5a 09 51 20 20 61 20 49
20 61 20 49 20 61 20 49 20 61 20 66 68 75 20 66
5a 66 09 51 20 69 51 20 67 09 53 20 09 51 20 20
61 20 49 20 61 20 49 20 61 20 49 20 61 20 68 73
61 69 64 5a 09 51 20 49 20 61 20 49 20 61 20 49
20 61 20 49 20 61 20 68 73 68 65 20 5a 09 51 20
49 20 61 20 49 20 61 20 49 20 61 20 49 20 61 20
68 41 6e 64 20 5a 09 51 20 54 5a 42 58 ff 62 ff

Some explanations on how it works:
Useful instructions are:

push imm8, push imm16 and push imm32, followed by pop generate constants. This can also generate zero (in ah) when pushing a byte (imm8).
and [ecx+32], ah - assuming ah = 0, this sets the byte to zero. It just so happens that the length of the output string is 32, so the code fills the buffer from end to beginning.
or [ecx+32], edx - assuming the output byte is set to zero, this copies edx (4 bytes) to output. I use a variant with dx instead of edx near the end of the buffer, because it should not write beyond the output buffer. The restriction on code makes it impossible to write single bytes this way!
imul edx, [ecx+32], whatever - this is the main scrambling idea. With enough entropy in [ecx+32] and the whatever number, it can generate any output. I use it to generate 2 or 3 bytes of needed values. Some complication is, when writing it to the output, it must do logical OR with whatever is already there. This sometimes made it necessary to zero the memory once again.
A variant of a jmp instruction is used to return. I chose it because its encoding is 0xff, which corresponds to a non-breaking space in codepage 437. A bit of a stretch on the rules, but otherwise I think the task is impossible...

Assembly source code, together with a C program which runs it (uses Visual Studio syntax):
#include <stdio.h>

__declspec(naked) void __fastcall doit(char* buf)
{
    __asm {
        push ' '
        pop eax

        dec ecx
        and [ecx+32], ah    // terminating 0 byte

        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah

        push 98
        pop edx
        or [ecx+32], edx
        imul dx, [ecx+32], 26183
        and [ecx+32], ah
        or [ecx+32], dx    // two bytes: [.']

        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push 'sih '
        pop edx
        or [ecx+32], edx    // 4 bytes: [ his]

        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push 538988610
        pop edx
        or [ecx+32], edx
        imul edx, [ecx+32], 544803081
        or [ecx+32], edx // 1 junk byte and 3 good bytes: (t's)

        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push 'aht '
        pop edx
        or [ecx+32], edx    // 4 bytes: [ tha]

        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        push 'tuB '
        pop edx
        or [ecx+32], edx    // 1 junk byte and 3 good bytes: [But]

        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push word ptr 8309
        pop dx
        or [ecx+32], dx
        imul edx, [ecx+32], 542312807
        or [ecx+32], edx    // 1 junk byte and 3 good bytes: [, ']

        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push 'dias'
        pop edx
        or [ecx+32], edx    // 4 bytes: [said]

        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push ' ehs'
        pop edx
        or [ecx+32], edx    // 4 bytes: [she ]

        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        dec ecx
        and [ecx+32], ah
        push ' dnA'
        pop edx
        or [ecx+32], edx    // 4 bytes: [And ]

        push esp
        pop edx
        inc edx

        pop eax
        jmp dword ptr[edx-1]
    }
}

int main()
{
    char buf[100];
    doit(buf);
    puts(buf);
}

This doesn't appear to be fully golfed to me. Surely you would save several bytes by using loops, rather than repeating a series of identical instructions. All of that dec ecx+and [ecx+32], ah stuff can be factored out.

You are welcome to try. This is the best I could do; I'd be happy to see a different approach. I decided to abandon the idea of loops when I saw that they require a negative jump offset. Maybe this restriction can be solved in a creative way - I don't know how.

@anatolyg It might come down to what your test environment's stance on self-modifying code is. Or its opinions on executing code you've built on the stack.

Doesn't char 0xff violate "No punctuation or non-alphabetic characters whatsoever"?

https://en.wikipedia.org/wiki/Code_page_437

The question says you have to print output, not return a string.  You should be able to tailcall a WinAPI DLL function, right?  Maybe the caller needs to pass a pointer to a print function (justification: makes this function nice and generic :)  Then you can push that pointer on the stack (below or above your own return address) and jmp to it?  Maybe we can construct a pointer in another register that lets us access [esp-4] with an actual addressing mode of jmp [ecx+32].

Or port to Linux (int 80h or syscall) or DOS (int 21h) or bootloader (BIOS or store directly to VGA RAM).  You would need self-modifying code to create the code bytes for int 0x80 (CD 80) or syscall (0F 05), or sysenter (0F 34).

PostScript, 889 874 837 835 bytes
currentflat string dup rrand
count dup count count mul mul xor count count mul count dup mul exch count
count copy count copy count copy count copy count copy
add and sub put print
and sub add put print
sub sub add put print
mul or xor put print
idiv xor add put print
or xor add put print
mod idiv add put print
mul or xor put print
idiv xor add put print
sub and add put print
or and add put print
sub sub add put print
pop add sub put print
mul or xor dup copy put print
mod mul sub put print
add or xor put print
idiv add add put print
add or add put print
put print
add or add put print
or xor add put print
sub and add put print
add or add put print
mod mul sub put print
idiv xor add put print
mul or xor put print
or xor add put print
or and add put print
idiv xor add put print
xor add sub put print
mod mul sub put print
quit

Try it online!
This uses 32 copies of the integers 89 25 20 6. All charcodes of the target string can be obtained with operations on those integers, in stack order: for example, 'A' (ASCII 65) is 89 - (25 & (20 + 6)). Many 4-tuples of integers have this property; this one was chosen because they’re particularly easy to generate.
currentflat string dup rrand

Flat defaults to 1, so this creates a string of length 1 (initialized to \0). dup is not a deep copy: it creates a second reference to the same string. rrand pushes the random seed, which defaults to 0. Stack is now ["\0", "\0", 0].
count dup count count mul mul xor

count pushes the number of items in the stack, so this computes 3 ^ (3 * (5 * 6)) = 89.
count count mul count dup mul exch count

4 * 5 = 20, 5 * 5 = 25, 6 = 6. The stack is now ["\0", "\0", 0, 89, 25, 20, 6].
count copy count copy count copy count copy count copy

Duplicate the entire stack, five times. Thus we end up with 32 copies of our initial 7-element stack. We only need 31 copies, since the target string is 31 characters long, but the extra copy doesn’t hurt.
add and sub put print

Compute a charcode from the top four integers, write it at index 0 of the string, then print the string.
quit

Suppresses the default prompt.

Ruby, 420 354 338  bytes
def P a
print String a
end
def Q a
p String a
end
class String
def inspect
putc sum size
q
end
end
def w
Q def hacked
end
rescue
end
P def And
end
w
P def she
end
w
P def said
end
Q def gadget
end rescue
w
def a
Q def afraid
end
rescue
end
a
P def But
end
w
P def that
end
a
putc String def s
end
w
P def his
end
Q def fierce
end rescue
a

Try it online!
In ascending order of jankiness:
Words starting with a capital letter can be printed by defining a class with that name and calling display within the class definition body.
Other words can be displayed by defining methods with that name, which returns a Symbol, then casting that to a String to remove the leading colon.
Other characters can be displayed by calling putc on their ASCII code. We can generate the appropriate numbers by reusing the String def trick to get a string, then taking the sum of its bytes using a modulus determined by its size. Unfortunately, we don't have any way to call methods on an object other than from within the class definition of that object, which makes it difficult to pass in arguments. So the final hack is to redefine String#inspect, which gets called implicitly when passing a String in to the p method, so that it calculates and outputs the appropriate character as a side effect, before raising an error so that p can't actually finish executing and print a newline. Then we need to rescue the error in the main code.
Edit: Jordan made the byte count much less, ahem, high with some clever control flow golfing, and I've cut a few more bytes by replacing raise with a one-letter nonexistent method call, which raises a NameError.
Edit 2: Noticed that with print String extracted into a method, it's cheaper to just use that with a method definition than to use the class definition trick, since methods are allowed to be title cased.

Beautiful.... I don't understand how sum size gets the sum modulo its size, but everything else checks out!

I put that a bit lazily, it's actually passing in the size of the string as the optional argument to the sum method.

What is the meaning of gadget, afraid, rescue, and fierce? Do the names do something, or would shorter ones suffice?

If I remember right, I was being playful and just looked for short English words that happened to give the numbers I needed, so probably they can be shortened. They're constrained because both the sum of the byte values and the length of the string matter.

><>, 233 122 bytes
cdacaabbglccgpcbfbbcaacdebbafbebbcebdbealbcpcdbcccdlcdpacbbalccpaalacpbcfafbaab

       g  sandBe
       o  Aviuth

Try it online!
This started as a golf of mbomb's answer, but I discovered a fundamental shift that saves a huge number of bytes, so I'm posting it as my own answer.
Generating non-alphabetic characters to output is done by repeatedly pushing values to the stack, then using l to push the stack length.  However, this does not need to be output immediately: using p, this character can be placed in any cell whose coordinates are both between 10 and 15 inclusive, to be retrieved later with g.  Similarly, alphabetic characters can be placed in the initial source code and read this way: since the highest non-alphabetic character code in the input is 46 (.), this means the stack does not need to be pushed higher than the 62 needed to store all 31 characters of the output.
Additionally, a v is placed in the code at column 7.  When the instruction pointer wraps around and hits that v, the sequence go is executed repeatedly to read from the pushed coordinates and output the corresponding characters.  Eventually, the stack becomes empty, and g terminates the program with an error.
The first 7 bytes of code are reused as both the first 7 and last 7 coordinates pushed.  Placing the v in column 9 would have theoretically saved two more bytes, but would have forced the characters Ainsv into a 2x2 square in the code, which is impossible.  An earlier version used column 15, but that required an extra line in the source code and ended up six bytes longer.

After further thought, I think I can make column 9 work by spending a byte on r to move the alignment wherever I want.  However, golfing this program hurts my brain a bit.

Stax, 133 95 84 79 bytes
AHHvcAvSVvNvXvsrDhYcvxvZLcHxVahyvxxZVvHVChAHHvVAHhvZAJyVahxvVvDhZyvxvAJZcASVAhL

Run and debug it

CJam, 262 bytes
  KAbScibCmhc  CZbsic          GmQYbsic
S CmfYmeibc    ImqmeKmhcel     AZbAbc
S CmfYmeibc    KAbScibCmhcel   ImqmeKmhAmhcel  GmQYbsic    KAZbYbbBbc
S CGmQbDbc     EYbTYtZbc       FYbGmQbcel      EYbGmQbcel
S EYbGmQbcel   ImqmeKmhcel     KAbScibCmhcel   EYbGmQbcel  CGmQbDbc    CmfYmeibc
S ImqmeKmhcel  ImqmeKmhAmhcel  CmfYmeibc       PYmhmeKmhc  CGmQbDbc

Try it online! Newlines are only shown here for clarity; each line represents a character.
Whew, this was fun. Restricting ourselves to alphabetic commands poses some interesting challenges:

Without { and }, there's virtually no opportunity for control flow (except f, which I didn't find an opportunity to use).
Without \, _, ;, or $, we have no means for stack manipulation.

This means that the main goal is going to be getting the relevant code points on the stack and then converting them to characters with c.
The problem is that we also lack most basic arithmetic commands, as well as integer literals. This is fine though, as the m namespace contains numerous advanced mathematical operations, and there are many variables predefined to useful numbers.
I ended up making heavy use of square roots (mQ and mq), the exponential function me, and base conversion (b), which can also be used to emulate multiplication ([X 0] Yb computes X*Y). In addition, sometimes it is easier to construct the uppercase codepoint, in which case we can use el (convert to lowercase) on the resulting character.
I'm still not satisfied with some of the longer ones. Oh well.
Explanation
This is a character-by-character explanation of the output. Before I start, here are a few short ways to make numbers:

0, 1, 2, 3 are contained in variables T, X, Y, Z respectively.
Numbers 10 through 20 are contained in variables A through K.
32 can be made using Sci (S pushes a string containing a space, c gets the first character of this string, and i converts that character to its code point). S is also used for spaces.
4 is given by GmQ (integer square root of 16).
5 is given by AZbYb (convert 10 to base 3, yielding [1 0 1], and convert the resulting array of numbers to base 2, yielding 5).
7 is given by Ymei (compute exp(2) and convert to integer).

A
K           - push 20                        | 20
 Ab         - convert to base 10             | [2 0]
   Scib     - convert from base 32           | 64
       Cmh  - hypot(TOS, 12)                 | 65.115
          c - round down and convert to char | 'A

n
C      - push 12            | 12
 Zb    - convert to base 3  | [1 1 0]
   s   - convert to string  | "110"
    i  - convert to integer | 110
     c - convert to char    | 'n

d
GmQ      - push 4             | 4
   Yb    - convert to base 2  | [1 0 0]
     s   - convert to string  | "100"
      i  - convert to integer | 100
       c - convert to char    | 'd

s
C         - push 12         | 12
 mf       - factors         | [2 2 3]
   Ymeib  - base 7          | 115
        c - convert to char | 's

h
I           - push 18                        | 18
 mq         - sqrt                           | 4.242
   me       - exp                            | 69.591
     Kmh    - hypot(TOS, 20)                 | 72.408
        c   - round down and convert to char | 'H
         el - lowercase                      | 'h

e
A      - push 10              | 10
 Zb    - convert to base 3    | [1 0 1]
   Ab  - convert from base 10 | 101
     c - convert to char      | 'c

a
KAbScibCmhc   - push 'A (see above) | 'A
           el - lowercase           | 'a

i
I              - push 18         | 18
 mq            - square root     | 4.242
   me          - exp             | 69.591
     Kmh       - hypot(TOS, 20)  | 72.408
        Amh    - hypot(TOS, 10)  | 73.095
           c   - convert to char | 'I
            el - lowercase       | 'i

,
K          - push 20              | 20
 AZbYbb    - convert to base 5    | [4 0]
       Bb  - convert from base 11 | 44
         c - convert to char      | ',

'
C        - push 12              | 12
 GmQb    - convert to base 4    | [3 0]
     Db  - convert from base 13 | 39
       c - convert to char      | ''

B
E         - push 14               | 14
 Yb       - convert to base 2     | [1 1 1 0]
   TYt    - replace elem 0 with 2 | [2 1 1 0]
      Zb  - convert from base 3   | 66
        c - convert to char       | 'B

u
F          - push 15             | 15
 Yb        - convert to base 2   | [1 1 1 1]
   GmQb    - convert from base 4 | 85
       c   - convert to char     | 'U
        el - lowercase           | 'u

t
E          - push 14             | 14
 Yb        - convert to base 2   | [1 1 1 0]
   GmQb    - convert from base 4 | 85
       c   - convert to char     | 'T
        el - lowercase           | 't

.
P          - push pi                        | 3.141
 Ymh       - hypot(TOS, 2)                  | 3.724
    me     - exp                            | 41.437
      Kmh  - hypot(TOS, 20)                 | 46.011
         c - round down and convert to char | '.

You can do base conversions with digits outside of the usual range for the base, such as HYbYCtYbc, HYbXAtYbc and HYbXBtYbc.

Deadfish~, 943 bytes
iiisdsiciiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiicddddddddddcddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddcdddddddddddddddddddddsddddddcdddddddddddcdddcdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddcdddddddddddddddddddddsddddddcddddddddddddddddddciiiiiiiicdddddcddddddddddddddddddddddddddddddddddddddddddddddddddddddddcddddddddddddciiiiiiiciiiiiiiiiiiiiiiiiiiiiiiiiiiciiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiicdcddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddcdddddddddddddddddddddsdddddcddddddddddddcdddddddciiiiiiiiiiiiiiiiiiicdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddcddddddddddddddddddddddddddddsddddddcdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddcddddddddddddddddddddddsiiiiciciiiiiiiiiicdddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddddcdddddddc

Try it online!
No loops allowed :(

evil, 198 bytes
aeeeannkhhwzuuuuuueaeuekwuuuuuuuuuunkhwzaeeeeehknwgueeywguuuuuuhhknnwuuuwhgwpwnngheeuwguuuuuwngwzaeeeaaaeeeeeewhhgwnguuuueewnngawpaawuwnngwpawhgwhhgeeuwpawhguuuueewpwhgwhgwawpwnngaaaaaaeeeewguuuueew

Try it online!
This was quite fun.

dc, 240 222 209 bytes
OOOOOziOOOOOOOOOOOOOOOOOOOOOOOOOOOzddddddzkdddzasBdzasAdzscdzdasCzsdOOlAxlAxPOBlBxdIlAxoPAdlBxddsrIlAxssPPOPlsIZlCxddspPOZlCxPPOPlrdZlCxPlsPlrPlcPPKPdZlBxdZlAxPOAZlAxdPIZlCxdPrPdPlpPlrdZlCxPPKPOPPlpPlsPOPldPKP

Try it online!
My first thought was the same as @seshoumara, just push enough stuff onto the stack to generate all of the characters' ASCII values. Then it occurred to me that since +, -, and * are single-character operators, I can just recreate them and have the ability to use arithmetic! Surely that would be smaller! And, I wouldn't be surprised if I'm able to golf off any more bytes, but for now... this convoluted approach has managed to tie the naïve(ish) one.
OOOOOziOOOOOOOOOOOOOOOOOOOOOOOOOOOzddddddzkdddzasBdzasAdzscdzdasCzsd is the part of the approach that's similar to @seshoumara's, but we only go up to 46, which is .. We do this because we need to go up to 45, -, and we also need a period in our string, so just going one further for the period is (I think) cheapest. Along the way, we store some values: 5, 32, 39 all come in handy later. 5 for utilitarian things, 32 and 39 for their ASCII values. Originally I did 1-5, but that was expensive, and I was able to simply avoid using 4; use Z (pop a value, push the number of digits it has) on a three-, two-, or one-digit number for those values. At 42, 43, and 45, we convert these to strings (*, +, and - respectively) and store them as macros (B, A, and C respectively). This means that without using the characters *+-, we can now use those operators.
From here we basically start generating the ASCII values using the power of math instead of sheer accumulation, storing some of the repeats along the way. 100, 105 and 115 come up enough that storing them (in registers or otherwise) made sense. Originally, I left the stack filled with 10s and used these to make 100s; it ended up saving bytes to fill the stack with 32s and use those as spaces later on. A slightly more readable version of the ASCII section: OOlAxlAxP OBlBxdIlAxoP AdlBxddsrIlAxssP P OP lsIZlCxddspP OZlCxP P OP lrdZlCxP lsP lrP lcP P KP dZlBxdZlAxP OAZlAxdP IZlCxdP rPdP lpP lrdZlCxP P KP OP P lpP lsP OP ldP KP.
Shaved off 18 bytes by: storing the number 5 as the input radix instead of a register; the number 32 as precision instead of a register; the number 115 as output radix instead of a register; then had to change KZ to IZ to generate 1s and OZ to KZ to generate 2s.
Shaved off 13 more bytes by flooding the stack with 32s; setting precision to 39; using stack manipulation to avoid storing 116; cutting out some double-work I accidentally left in.

+1 Nice idea to use a to recreate those operators, then to call them with x. This shows the data-is-code behavior of dc. When I have time I'll apply your latest trick of storing data into parameters, instead of registers. Do you think we could get an even shorter dc solution by abusing the way P works to print more letters at a time if we're lucky to input the needed huge number using hex only?

@seshoumara It's possible, though my attempts at doing that with other challenges suggests that it's unlikely simply because those values become large quickly. Just to get the first two letters, 'An', we need to P 16750 or 0x416E. If we just happened to get lucky and one of the substrings was comprised exclusively of values A-F, then that might get us a shortcut. That would be some luck, though! Otherwise we'd be either inputting large numbers somehow, coming up with them somehow, or doing a lot of adding and multiplying by 256. Which seems... chunkier than a bunch of Ps.

MATL, 187 158 bytes
IEWQKEtqhpEqqKQHhthpKWEyQKWEqqYqQQXJwtQQQwKWEJKQthpYqKQHhthptQQQQQwIIhKQhpqKWEIWKQhpqIEWQQJQQtqKWEyIIhIhpYqQXJyKQthpYqwIWKQhpqyqKWEyJtQwhhPIIhKQhpQIWKQhpqvlec

Try it online!
More readable version: Try it online! Manual attempt at building the string. There's probably plenty of room for golfing by cutting the string into convenient chuncks, using P and h to flip and build a string. I hope somebody will take up the challenge to outgolf me. The main challenge is that you cannot use + or -, so basic arithmetic is often not possible.
Highlights:

KQthpYq: the 25th (KQthp) prime Yq is 97, corresponding to the letter a. The letter s (115) is generated similarly from 113, the 30th prime number. It is then extensively re-used in clipboard J.
his is shortened by storing the h from before in clipboard J.  Since that previously stored s, we build his in reverse so we can still retrieve the recent s using y, and flip it after using P.
A large number of bytes saved thanks to Luis Mendo (mostly by changing a bunch of h to vle)

Huh - I really thought using v would mess with the h I had earlier. Guess I should've that instead of just assuming. Also, thanks for the heads-up, overzealous deleting happened. I'm curious to see if you can do better than this...

I'd like to try, but it will have to wait. Answering this seems to take a lot of time!

@LuisMendo Yup. Took me an hour or so to make this, and this only includes very local optimisation. I'm sure one can do better with a bit more of a global look at things.

My attempt. I didn't look at yours, so the approaches are hopefully different

@LuisMendo I completely forgot that U means ^2, that could've saved me quite a few bytes...

Japt, 87 bytes
Bs g caA
HzG
Ts caV
iWisiiihiSisiUitiaihitiSitiuibu iUiSiWcaV idiiiaisiSieihisiSidiniau

Try it
Explanation
The first line generates the ' and assigns it to variable U.
Bs g caA
B                            :11
 s                           :To string
   g                         :First character
     c                       :Character code
      a                      :  Absolute difference with
       A                     :  10

The second line assigns 2 to variable V.
HzG
H                            :32
 z                           :Floor divided by
  G                          :16

The third line generates the . and assigns it to variable W.
Ts caV
Ts                           :Convert 0 to a string
   caV                       :Absolute difference of its charcode with V (2)

The last line, then, builds the string one character at a time in reverse.
iW...ibu ...iWcaV ...iau
iW                           :Start by prepending W (.) to U (')
  ...                        :Each i prepends the literal character that follows it to the string, with S being space and U being "'"
     ibu                     :As B is the constant for 11 and it can't be quoted, here i prepends "b" to the string and u uppercases it
         ...                 :As above, each i is prepending the character/constant that follows it to the string
            iWcaV            :Gets the absolute difference of the charcode of W (.) and V (2) to get the "," and prepends that
                  ...        :Some more literals
                     iau     :And, finally, the same trick is used for the "A" as was for the "B", as A is the constant for 10

Nice solution. You can save a byte by replacing the first line with Qc dGaB

MATL, 118 bytes
KEUQtVQsQKBFZAXHKUqyyhsXJyyhdQtQQQwOJIUEyhdtKEhsHKQYqEEKQyywhdXJKEUQQHKUhstQwOytHKhsKYqyhdbJyqOHKhstQHKUqhsKUqYqqJvlec

Try it online!
More readable version (each line corresponds to one character, except stack-rearranging operations).
Explanation
The program produces the code points of the required characters, as independent numbers. At the end all those numbers are concatenated into a column vector, reshaped as a row vector, and converted to characters. The result is implicitly displayed.
Some of the tricks used:

Most code points below 32 are displayed as space. So 0 is used for most spaces, because it only costs a byte (O).
For the first space, however, 15 is used (generated as KUq), because it can then be reused by adding it to 100 (char d) to give 115 (s). On another occasion 5 is used for space (generated as KQ), so it can be later subtracted from 44 (,) to give 39 (').
Clipboard J is used for storing characters that will be repeated: first s, then '. Similarly, clipboard H stores 100, which is useful for d and for generating other characters.
Extensive use is made of functions Q (add 1), q (subtract 1), E (multiply by 2) and U (square), together with the predefined literals in clipboards I (3)  and K (4).
Arbitrary addition and subtraction are done by concatenating into a vector (h) and computing its sum (s) or consecutive differences ( d).
100 (d) is generated as 4 in binary interpreted as a decimal number.
110 (n)  is obtained from 65 (A) by converting to string ('65': code points [54 53]), adding 1 to the code points ([55 54]), summing them together and adding 1.
The order in which numbers are generated is sometimes altered for convenience; and they then are reordered by stack-rearranging functions: swap (w), bubble up b).

Very nice! Clever touch using O instead of KWE for spaces. And you confirmed my suspicion that it's probably better to sacrifice another clipboard H.

Red, 272 bytes
prin quote And prin sp prin quote she prin sp prin quote said prin comma prin sp prin subtract to sp mold quote G sp prin quote But prin sp prin quote that prin subtract to sp mold quote G sp prin quote s prin sp prin quote his prin dot prin subtract to sp mold quote G sp

If the double quotes are necessary:
Red, 344 bytes
prin subtract to sp mold quote B sp prin quote And prin sp prin quote she prin sp prin quote said prin comma prin sp prin subtract to sp mold quote G sp prin quote But prin sp prin quote that prin subtract to sp mold quote G sp prin quote s prin sp prin quote his prin dot prin subtract to sp mold quote G sp prin subtract to sp mold quote B sp

Doesn't work in TIO but works in the Red interpreter.

Explanation:
The words are trivial - I prin them (print with no newline) as literals with quote. Red has a built-in word for space - sp, as well as comma and dot. " and ' more interesting: I prin them by subtracting a space from B and G respectively, starting from a literal B and G, converting them first to string with mold and then to character (in order to use subtraction on them) wtih to sp (Red has conversion by prototype - convert the string to the type of sp, which is character).

The question was clarified; they removed the interesting part from it.

@anatolyg Thanks, I stil need it for ', so the 272 bytes solution is the same.

Forth (gforth), 351
CHAR A DUP EMIT
CHAR n EMIT
CHAR d EMIT
SPACE
CHAR s DUP EMIT
CHAR h EMIT
CHAR e EMIT
SPACE
EMIT
CHAR a EMIT
CHAR i EMIT
CHAR d EMIT
DUP CHAR m XOR EMIT
SPACE
CHAR f XOR DUP EMIT
CHAR B EMIT
CHAR u EMIT
CHAR t DUP EMIT
SPACE
DUP EMIT
CHAR h EMIT
CHAR a EMIT
EMIT
DUP EMIT
CHAR s EMIT
SPACE
CHAR h EMIT
CHAR i DUP EMIT
CHAR s EMIT
CHAR G XOR EMIT
EMIT

Too bad I can't redefine CHAR or EMIT to one-letter words, since that would require using either : and ; (e.g. : C CHAR ;) or ' (e.g ' CHAR ALIAS C)
In fact, if I could define words, I could do : P CHAR EMIT ; and then do P x to print x. Oh well.
I can't even create a buffer, write that char sequence to there and then use that as the input, since writing to memory requires using ! or C!

AlphaBeta, 180 177 175 163 bytes
cccaaggtFgDILrFigggDLjDLCLigggggDLjhDLhhhDLCLiggggDLjjggDLihhDLhhhhhDLcaaCLdbbCLcbbbCLHgDLiiiiigDLhDLdaaaCLDLjhhDLjgggDLiihDLcbbbCLhDLdaaaCLjhDLgDLiDLcaaaaCLdaaaCL

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WIP

Pepe, 266 Bytes
I keep the r stack empty, and have 's' on the R stack
reeEeeeeeE reeEEeEEEe reeEEeeEee reEe REeEEEeeEE Reee reeEEeEeee reeEEeeEeE reEe Reee reeEEeeeeE reeEEeEeeE reeEEeeEee reeeEeEEee reEe reeeEeeeEe reeEeeeeEe reeEEEeEeE reeEEEeEee reEe reeEEEeEee reeEEeEeee reeEEeeeeE reeEEEeEee reeeEeeEEE Reee reEe reeEEeEeee reeEEeEeeE Reee reeeEeEEEe reeeEeeeEe

This isn't on TIO, but you can try it here

Befunge-98 (FBBI), 125 124 121 bytes
wab

And she said   But that s his

wakekekaayyeapwayyaayybyapayybyapcyabcyaayycayyba
a

b
wayapapoq

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Outputs to a file named \n (a single newline). Thanks to Jo King for his script.
Output includes 10 trailing newlines.
For only one trailing newline, +1 byte by changing the following line:
wakekekaayyeapwayydayybyapayybyapycyabcyaayycayyba
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Explanation:
The instruction pointer moves as follows:
The program puts the non-alphabetic characters into position, before outputting that line to a file.
Befunge-98 includes instructions a...f, which push their corresponding hexadecimal value to the stack. To generate other numbers, it passes those values to y ("Get SysInfo") as arguments to obtain:
10  y-position
11  x-position
12  y-velocity (= 0)
13  x-velocity (= 1)

23* stack size

By placing the bulk of the code at y=23, ayy can be used for repeated access to the stack size, which is then used to generate character codes.

Isn't only one trailing newline allowed?

The post states "Whitespace is completely allowed". I think it's cool to abuse this wording to justify trailing newlines!

dc, 240 bytes
The main idea is to continuously grow the stack by 1 (K), saving (sX) the size of the stack (z) into custom registers when it matches each unique ASCII code. Printing (P) is done throughout.
KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKzsSKKzsQKKKKKzsqKKKKKzsCKKzsDKKKKKKKKKKKKKKKKKKKzPKzsBKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKzsaKKKzsdKzseKKKzshKzsiKKKKKzPldPlSPKKKKKzsszPlhPlePlSPlsPlaPliPldPlCPlSPlqPlBPKzdstzPdPlSPdPlhPlaPPlqPlsdPlSPlhPliPPlDPlqP

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I did some optimization, like not saving a letter if it is not used after that, like duplicating (d) a letter, e.g. t, on the stack to save a byte, since recall (lX) is 2 bytes.

I came up with a dc solution that uses arithmetic by creating macros for +, -, and *. It’s 262 bytes. I’ll keep trying to optimize it to try to get competitive, but I must say I’m disappointed it’s so much fatter than the (relatively) naïve solution.

@brhfl Yeah, dc gets verbose pretty fast. Even so, I would very much like to see your solution, so post it! In the meantime, I'm also thinking of either golfing the current method more, using a program, or coming up with another idea for dc.

I managed to tie 240! And I wouldn't be surprised if I was able to golf off one or two more bytes, but... it's a much more obtuse approach for little/no gain. Anyway, I posted it down there somewhere...

I got down to 222 by replacing some of my most common registers w/ input/output radix and precision; saves one byte on the store and one byte on every load... Since numbers are immaterial, it doesn't affect anything... You might be able to use this to your advantage as well!

80186+ machine code, MS-DOS .COM format, 822 787 bytes
Only tabs and spaces are used in addition to letters. Given that most of the opcodes in the allowed range are certain increments, decrements, pushes, pops, and register-indirect ANDs and ORs, in addition to IMUL, I make use of the fact that the stack register wraps around when it hits the end of the segment to modify the code in reverse! 80186+ assembly is required because I am pushing immediate values.
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXjhX   GXkOXYIQSX GXjdX    GXkOXwIIIIIIIIIQhhihs kOXeQh ihutSXH    GXHHHHHH GXSX GYkOXDAAAQSX GXjGX    GXkOXtQhidhsahe hshhd hAnSX GXjTX   GXkOXdIIIQkOXgAQSX GXHHHHHHHHHHHHHHHHHHHHH  GXSX GYkOXbAAAAAAAAAAAAAAQhhlh  Xhh qM

Annotated source (TASM format):
IDEAL
P186

MODEL   TINY
CODESEG
ORG 100H

MAIN:
REPT 582
    POP AX  ; Set up stack to end of string
ENDM

    PUSH 68H
    POP AX
    OR [BX+58H],AX
    IMUL CX,[BX+58H],59H ; 68H*59H=2428H
    DEC CX ; -1=2427H
    PUSH CX

    PUSH BX
    POP AX
    AND [BX+58H],AL
    PUSH 64H
    POP AX
    OR [BX+58H],AX
    IMUL CX,[BX+58H],77H ; 64H*77H=2E7CH
REPT 9
    DEC CX ; -9=2E73H
ENDM
    PUSH CX

    PUSH 6968H
    PUSH 2073H

    IMUL CX,[BX+58H],65H ; 64H*65H=2774H
    PUSH CX

    PUSH 6920H
    PUSH 7475H

    PUSH BX
    POP AX
    DEC AX
    OR [BX+58H],AX ; FFFFH
REPT 6
    DEC AX
ENDM
    AND [BX+58H],AL ; FFF9H
    PUSH BX
    POP AX
    AND [BX+59H],AL ; 00F9H
    IMUL CX,[BX+58H],44H ; 0F9H*44H=4224H
REPT 3
    INC CX ; +3=4227H
ENDM
    PUSH CX

    PUSH BX
    POP AX
    AND [BX+58H],AL
    PUSH 47H
    POP AX
    OR [BX+58H],AX
    IMUL CX,[BX+58H],74H ; 47H*74H=202CH
    PUSH CX

    PUSH 6469H
    PUSH 6173H
    PUSH 2065H
    PUSH 6873H
    PUSH 2064H
    PUSH 6E41H

;; CODE STARTS:
;; PUSH 0909H
;; POP AX
;; PUSH 046CH
;; POP DX
;; INT 21H
;; INT 20H

    PUSH BX
    POP AX
    AND [BX+58H],AL
    PUSH 54H
    POP AX
    OR [BX+58H],AX
    IMUL CX,[BX+58H],64H ; 54H*64H=20D0H
REPT 3
    DEC CX ; -3=20CDH
ENDM
    PUSH CX

    IMUL CX,[BX+58H],67H ; 54H*67H=21CCH
    INC CX ; 21CDH
    PUSH CX

    PUSH BX
    POP AX
    AND [BX+58H],AL
REPT 21
    DEC AX
ENDM
    OR [BX+58H],AX ; 0FFEBH
    PUSH BX
    POP AX
    AND [BX+59H],AL ; 0EBH
    IMUL CX,[BX+58H],62H ; 0EBH*62H=59F6H
REPT 14
    INC CX ; +14=5A04H
ENDM
    PUSH CX

    PUSH 6C68H
    PUSH 5809H
    PUSH 0968H

    JNO $+4FH

END MAIN
ENDS

This isn't assembly, this is machine code...

@Artelius Fair enough. I've updated the description.

Vyxal, 112 bytes
kAhkHLnikhtnukWhkgTninukahkfuTikhtkTtnkekbhnuTdikWstnkWstkWhkahkWstkeunkWhkfuTiukekTy_tWnuNIVunkiTdiVkekHLTCVuIj

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This was created by hand and it's a mess that I won't explain.

Pyth, 210 bytes
pChyCdpCyhyhlGpCyytlGpdpChFhTyylGpCyylGpChyytlGpdpChFhTyylGpCtytytlGpChyylGpCyytlGpCyyhTpdpCtyyTpCyhCdpCtFyyCdhTpCyhFlGCdpdpCyhFlGCdpCyylGpCtytytlGpCyhFlGCdpCtyyTpChFhTyylGpdpCyylGpChyylGpChFhTyylGpCyhyhTpCtyyT

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I found a few numbers which could be expressed using only letters (like T = 10, Z = 0, lG = length(alphabet) = 26, Cd = charcode(space) = 32), and a few functions that could be performed using only letters (like t = decrement, h = increment, hF = repeated application of increment = addition), and then just ran a brute force search to find the shortest combinations of those functions and numbers that resulted in each of the letters I needed.

Japt, 74 bytes
HdEz
iHdE isiiihiSisiUitiaihitiSitiuibu iUiSiHdC idiiiaisiSieihisiSidiniau

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16-bit x86 assembly code, 665 bytes (the binary is alphabetic, not the source)
I somehow forgot about the rule allowing whitespace.  Surely the code can be golfed as a result.
Bytecode:
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXsBFVKZPFFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPFPXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXkLAFQQZJJJRkDCGPLXDPDJRkDBEPZJJRLZDRDZAAAQPLYDQDYXXDQhishZhDRDhZsDQDhaththRDhuthZBDQDRhidhsaRDhhehZsDRDhndhZADTZPiDEFY
Source:
    db    63 dup (58h) ;pop ax
    jnb   label1
    dw    5646h      ;magic #1
    dw    5a4bh      ;magic #2
    dw    4650h      ;magic #3
    dw    (42h-6)/2 dup ("PF")

label1:
    db    416 dup (58h) ;more pop ax
    imul  cx,[si+41h],46h ;cl=24h (string sentinel)
    push  cx         ;push string sentinel
    push  cx
    pop   dx         ;dl=24h
    dec   dx
    dec   dx
    dec   dx         ;dl=21h
    push  dx         ;save for later
    imul  ax,[si+43h],47h ;al=0CDh
    push  ax         ;push xxCDh
    dec   sp         ;insert xx
    pop   ax         ;ah=0CDh
    inc   sp         ;discard xx
    push  ax         ;push 0CDxx
    inc   sp         ;discard xx
    dec   dx         ;cl=20h (space)
    push  dx
    imul  ax,[si+42h],45h ;al=2Eh (dot)
    push  ax
    pop   dx         ;dl=2Eh
    dec   dx
    dec   dx         ;dl=2Ch (comma)
    push  dx         ;push xx2Ch
    dec   sp         ;insert xx
    pop   dx         ;dl=2Ch
    inc   sp         ;discard xx
    push  dx         ;push 2Cxxh
    inc   sp         ;discard xx
    pop   dx         ;dx=202Ch
    inc   cx
    inc   cx
    inc   cx         ;cl=27h (quote)
    push  cx         ;push xx27h
    push  ax         ;push xx2Eh
    dec   sp         ;insert xx
    pop   cx         ;ch=2Eh
    inc   sp         ;discard xx
    push  cx         ;push 2Exxh
    inc   sp         ;discard xx
    pop   cx         ;cx=272Eh
    pop   ax         ;discard xxxx
    pop   ax         ;ax=0CD21h
    inc   sp         ;discard xx
    push  cx         ;push ".'"
    push  7369h      ;push "is"
    push  685ah      ;push "h"+xx
    inc   sp         ;discard xx
    push  dx         ;" "+xx
    inc   sp         ;discard xx
    push  735ah      ;push "s"+xx
    inc   sp         ;discard xx
    push  cx         ;push "'"+xx
    inc   sp         ;discard xx
    push  7461h      ;push "at"
    push  6874h      ;push "th"
    push  dx         ;push " "+xx
    inc   sp         ;discard xx
    push  7475h      ;push "ut"
    push  425ah      ;push "B"+xx
    inc   sp         ;discard xx
    push  cx         ;push "'"+xx
    inc   sp         ;discard xx
    push  dx         ;push ", "+xx
    push  6469h      ;push "id"
    push  6173h      ;push "sa"
    push  dx         ;push " "+xx
    inc   sp         ;discard xx
    push  6568h      ;push "he"
    push  735ah      ;push "s"+xx
    inc   sp         ;discard xx
    push  dx         ;push " "+xx
    inc   sp         ;discard xx
    push  646eh      ;push "nd"
    push  415ah      ;push "A"+xx
    inc   sp         ;discard xx
    push  sp
    pop   dx         ;dx=sp
    push  ax
    imul  ax,[si+45h],5946h ;ah=09h
It works this way:

moves the stack pointer to the end of the code, via POP AX (can't POP SP because it's not alphabetic);
constructs the instruction to dispatch a DOS call (algorithmically because it's not alphabetic);
constructs the non-alphabetic characters;
places the string on the stack;
places the dispatch instruction on the stack at the exact end of the code, so that execution flows directly to that instruction;
constructs the instruction to print a string;
displays the string and them promptly crashes. :-/ (A graceful exit would require more code)

05AB1E, 145 121 109 bytes
TnVYTOUXZxNJTZXTZZZXYTYxMNYTTMNNNYRYNNYNNNNYTXZMNNYxXRZZZXTYXRZZXZYxTRMRNXRMRMXNNYxYMRNTYxTRYNTZMYXRXXXTZJCXB

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Husk, 408 bytes
mcJLQtDDQDsisNeJDDDDDisNeJDDDDDisNeJLDQQDDsisNeLtQtDDQDsisNLDDtDtDtDDDsisNJLtQDtDDDsisNeLtDDtDtDDDDsisNLttttQDtDDDsisNJLDDtDtDtDDDsisNeJLtDtDtDtDtDDDsisNeLtDDtDtDDDDsisNLQDtDDDsisNJDDDDDisNeDDiDsisNLtDDQDDsisNJLtDDQDDsisNeJLDDtDtDDDDsisNeJLDDtDtDDDDsisNeLtttPtDQDsisNDDDDDisNJLtDtDtDtDtDDDsisNeLtQDtDDDsisNLDDtDtDDDDsisNJLQDtDDDsisNeJDDDDDisNeLtDDtDtDDDDsisNLtQDtDDDsisNJLDtPDDsisNeLtDDtDtDDDDsisNLtDDQDDsisN

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Well, that was an adventure.
The code was generated with this Python script. Not sure it's optimal, but it's good enough for me.
Explanation
So, limiting Husk to using only letters turned out to be a real pain... My first idea was to use s to convert a value to string, use t (drop first element) and D (double) to change the length of this string, then convert it to a number by computing its length with L: I could do this to generate the ASCII code of each character in the output string, and the convert them to characters with c.
That was complex enough for my tastes, but then I realized that all the builtins associated to letters in Husk are functions (which cannot be converted to string), except for N... which is the infinite list of natural numbers, good luck getting the length of that. Fortunately, i can take a string and return the first integer found in that string, which for N would be 1, so with sisN I can get my string of length 1 and I can manipulate that to get all other lengths.
I wrote a python script to bruteforce the shortest way to build strings of every needed length starting from sisN and using t, D, P (permutations, works like a factorial), and Q (substrings, works like n*(n+1)/2). I also realized that some values could be built more efficiently by working directly on numbers rather than strings, so isN is 1, DisN is 2, iDsisN is 11 (double the string "1" and read that as a number) and so on... those values were not too many, so I just hardcoded them as alternatives and kept for each the shortest between the hardcoded string and the bruteforced one.
Only after I did all that I realized that I didn't have a way to combine different characters into an output string. Strings are just lists of characters, and e can build a list of two arguments, but to build longer lists I would need :...
This time I was saved by J, join, which (among other things) can take a value and a list and insert that value between each pair in the list. If I want to concatenate three strings a b c I can do it with Jbeac. The script does this recursively starting from the desired output by choosing b as a single central element and a and c as the two halves separated by b. This was my first attempt an miraculously it worked (if the length of the string was different I could have to join only two strings together, but this doesn't happen here). I have tried other ways to compose the string, but this remains the shortest working result I could produce.

Pip, 102 bytes
ndzshezsaidyutzthatxszhiswxPB AoADtYSTfYDQUWyRMsYyH U UoEoOPOAZkPB CxPBPOAZyRDQzsRDQzkRDQzCxRDQzPIRMXD

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Explanation
Eagle-eyed readers will have noticed that the code begins with all the lowercase letters from the target string, plus a few others sprinkled in. We'll get back to that in a minute. For now, Pip treats them as no-ops, except the last one:
xPB AoADt
    Ao     ; ASCII code of "1" (49)
      ADt  ; Absolute difference with 10 (39)
xPB        ; Push this result onto variable x (initially "")

This sets x to 39, the ASCII code of '. (Trick borrowed from Shaggy's Japt answer.)
YSTf
   f  ; The main function
 ST   ; Stringify
Y     ; Yank into y variable

Stringifying the main function formats the code and wraps it in curly braces. This is the result:
"{n; d; z; s; h; e; z; s; a; i; d; y; u; t; z; t; h; a; t; x; s; z; h; i; s; w; x PB ((A o) AD t); YP (ST f); Y (DQ (UW (y RM s))); Y (y H ((U (U o)) E o)); O (PO AZ); (k PB (C x)) PB (PO AZ); (((y R (DQ z); s) R (DQ z); k) R (DQ z); (C x)) R (DQ z); (PI RM XD)}"

From here, we can extract the leading run of lowercase letters:
YDQUWyRMsYyH U UoEo
     yRMs            ; y with spaces removed
   UW                ; Unweave into two every-other-character strings
 DQ                  ; Dequeue the last one, which has the letters in it
Y                    ; Yank that string into y
             U Uo    ; Increment o twice (was originally 1, now is 3)
                 Eo  ; Take that value to the power of o (27)
          yH         ; Get that many characters from the start of y
         Y           ; Yank that string into y

Now y is "ndzshezsaidyutzthatxszhiswx".
OPOAZ
   AZ  ; Uppercase alphabet
 PO    ; Pop the first character ("A")
O      ; Output it without a newline

Outputting the initial A separately is cheaper than making it part of the string.
kPB CxPBPOAZ
k             ; Take k (initially ", ")
 PB           ; Push this onto it:
    Cx        ;  Char(39) ("'")
      PB      ; Then push this onto it:
        POAZ  ;  Pop the first character from AZ ("B")

Finally:
yRDQzsRDQzkRDQzCxRDQzPIRMXD
                             ; z is preinitialized to the lowercase alphabet
yR                           ; In y, replace
  DQz                        ;  character dequeued from z ("z")
     s                       ;  with space
      R                      ; Replace
       DQz                   ;  character dequeued from z ("y")
          k                  ;  with k (", 'B")
           R                 ; Replace
            DQz              ;  character dequeued from z ("x")
               Cx            ;  with char(39) ("'")
                 R           ; Replace
                  DQz        ;  character dequeued from z ("w")
                             ;  with:
                     PI      ;   Pi
                         XD  ;   with digits
                       RM    ;   removed (".")

This expression gives us the rest of our target string, which is autoprinted.

BRASCA, 213 bytes
DoluRlugoHoEoDuRxluxluxMgoldUxgollUxgoEoDuRxluxluxMgoholDUmxMgoHoduxduxgoEoEuxhuxgoDuxDuxgoluxluxhumxMgoluxluxDuxgoEoluxluxDuxgoldUxgoholuxluxDuxgoEuxhuxgoDuRxluxluxMgoEoldUxgolDUmxMgoDuRxluxluxMgoduxDuxgoEuxhuxgo

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Explanation
At first I wanted to port the ><> answer, before realizing that stack length wont work with only alphabetical characters. That's when I remembered that BRASCA has some nifty things ><> does not have: The g and u operators!
Using the lLeEdDhHK built-ins for pushing numbers, u to seperate them into individual digits, R, m and M to rotate and shift the stack, x to discard digits, and then g to put them back together I can create and output any string using only the alphabet.
Breakdown
    DoluRlugoHoEoDuRxluxluxMgoldUxgollUxgoEoDuRxluxluxMgoholDUmxMgoHoduxduxgoEoEuxhuxgoDuxDuxgoluxluxhumxMgoluxluxDuxgoEoluxluxDuxgoldUxgoholuxluxDuxgoEuxhuxgoDuRxluxluxMgoEoldUxgolDUmxMgoDuRxluxluxMgoduxDuxgoEuxhuxgo
 +-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
A|  Do
n|    luRlugo
d|           Ho
 |             Eo
s|               DuRxluxluxMgo
h|                            ldUxgo
e|                                  llUxgo
 |                                        Eo
s|                                          DuRxluxluxMgo
a|                                                       ho
i|                                                         lDUmxMgo
d|                                                                 Ho
,|                                                                   duxduxgo
 |                                                                           Eo
'|                                                                             Euxhuxgo
B|                                                                                DuxDuxgo
u|                                                                                luxluxhumxMgo
t|                                                                                luxluxDuxgo
 |                                                                                Eo
t|                                                                                luxluxDuxgo
h|                                                                                ldUxgo
a|                                                                                ho
t|                                                                                luxluxDuxgo
'|                                                                                Euxhuxgo
s|                                                                                DuRxluxluxMgo
 |                                                                                Eo
h|                                                                                ldUxgo
i|                                                                                lDUmxMgo
s|                                                                                DuRxluxluxMgo
.|                                                                                duxDuxgo
'|                                                                                Euxhuxgo

Actually, 2679 2618 2561 bytes
 PPuPuPDDc uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuc buuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuck D cHj

Try it online!
I'll golf each character one by one. So far I've golfed up to A.
Explanation
I'll golf pushing the numbers over time, so I'll only explain k D cHj for now.
k       # Put all the characters into the list
  D cH  # Push the empty string
      j # Join the list using the empty string

80186 machine code + DOS, 91 bytes
Text version:
hm  j   j   PPjzjzjgaAAA    JSJJ    RU  Sq  ReAA    JdJJJ   RfiJElK JEiS GtI And she said   But that s his

Text version, with tabs (code 9) replaced by 9 and spaces (code 32) replaced by *:
hm9j9j9PPjzjzjgaAAA9JSJJ9RU9Sq9ReAA9JdJJJ9RfiJElK9JEiS*GtI*And*she*said***But*that*s*his***

Hexdump:
68 6D 09 6A 09 6A 09 50 50 6A 7A 6A 7A 6A 67 61
41 41 41 09 4A 53 4A 4A 09 52 55 09 53 71 09 52
65 41 41 09 4A 64 4A 4A 4A 09 52 66 69 4A 45 6C
4B 09 4A 45 69 53 20 47 74 49 20 41 6E 64 20 73
68 65 20 73 61 69 64 20 20 20 42 75 74 20 74 68
61 74 20 73 20 68 69 73 20 20 20

The machine code appears in a file with extension .com. When I run it, it prints the required message and then hangs (executing random data).
High-level explanation on what it does:

Initializes registers with constant values
Replaces spaces in the message by the required special symbols (,'.$)
Patches the code to generate the int 21 instruction, which prints the message
Calls DOS

Assembly code (can be compiled with tasm):
my_bp equ 7ah
my_si equ 7ah
my_di equ 67h
my_msg equ 13bh
    .model tiny
    .code
    .startup
    .186
    org 100h
    push 96dh   ; ax (ah = 0; al = don't care, but see below)
    push 9      ; cx
    push 9      ; dx
    push ax     ; bx = don't care
    push ax     ; don't care
    push my_bp
    push my_si
    push my_di
    popa
    inc cx
    inc cx
    inc cx
    or [bp+si+my_msg-my_bp-my_si+12], cx ; ,
    dec dx
    dec dx
    or [bp+si+my_msg-my_bp-my_si+14], dx ; '
    or [bp+di+my_msg-my_bp-my_di+23], dx ; '
    or [bp+si+my_msg-my_bp-my_si+30], dx ; '
    inc cx
    inc cx
    or [bp+si+my_msg-my_bp-my_si+29], cx ; .
    dec dx
    dec dx
    dec dx
    or [bp+si+my_msg-my_bp-my_si+31], dx ; $

    ; 0x2049 * 0x4b6c = 0x98301cc
    ; So this sets cx to 1cc (a temporary constant used to patch code)
    imul cx, [bp+si+my_msg-my_bp-my_si-2], 4b6ch
    ; 0x1cc | 0x2049 = 0x21cd (the instruction which calls DOS int 21)
    ; Here ah = 9 ("print" mode)
    or [bp+si+my_msg-my_bp-my_si-2], cx

    ; At address 101, there is the constant 96d, which was loaded into ax
    ; 0x96d * 0x7447 = 0x448013b
    ; So the following sets dx to 13b (adddress of the message)
    imul dx, [bp+di+101h-my_bp-my_di], 7447h

int21:
    dw 2049h

    db 'And she said   But that s his   '
    end

It uses the popa instruction to pop all registers, because regular pop cannot fill all needed registers (e.g. pop di is a forbidden opcode).
Addresses of bytes to patch are in the range 0x100...0x160. By luck, they can be represented as a sum of 3 bytes with allowed values:

0x7a in bp
0x7a or 0x67 in si or di
Immediate value

Patching of bytes in the message works by doing logical OR on 0x20 (space character) and a small constant (4, 7, 12 or 14). The small constant is obtained by initializing cx and dx to 9 (tab character) and doing INC or DEC as needed.
Patching of code uses the IMUL instruction. I found the needed 16-bit constants to multiply using brute-force search.
Finally, the address of the message (0x13b) is obtained by multiplication. To save space, I took one of the constants from one of the instructions, which contains an immediate value 0x96d. Here the 9 part chooses a DOS print function, and the 6d part is a free parameter. It turns out that 6d is the only possibility which can give 0x13b after multiplication.
Disassembly of the code part:
06BA:0100 686D09            PUSH    096D
06BA:0103 6A09              PUSH    +09
06BA:0105 6A09              PUSH    +09
06BA:0107 50                PUSH    AX
06BA:0108 50                PUSH    AX
06BA:0109 6A7A              PUSH    +7A
06BA:010B 6A7A              PUSH    +7A
06BA:010D 6A67              PUSH    +67
06BA:010F 61                POPA
06BA:0110 41                INC     CX
06BA:0111 41                INC     CX
06BA:0112 41                INC     CX
06BA:0113 094A53            OR      [BP+SI+53],CX
06BA:0116 4A                DEC     DX
06BA:0117 4A                DEC     DX
06BA:0118 095255            OR      [BP+SI+55],DX
06BA:011B 095371            OR      [BP+DI+71],DX
06BA:011E 095265            OR      [BP+SI+65],DX
06BA:0121 41                INC     CX
06BA:0122 41                INC     CX
06BA:0123 094A64            OR      [BP+SI+64],CX
06BA:0126 4A                DEC     DX
06BA:0127 4A                DEC     DX
06BA:0128 4A                DEC     DX
06BA:0129 095266            OR      [BP+SI+66],DX
06BA:012C 694A456C4B        IMUL    CX,[BP+SI+45],4B6C
06BA:0131 094A45            OR      [BP+SI+45],CX
06BA:0134 6953204774        IMUL    DX,[BP+DI+20],7447
06BA:0139 CD21              INT     21 (after the code patches itself)

Fun fact: Normally, I would use offset message instead of the hard-coded 13bh, but in this case, because at the time of parsing its address is unknown, tasm generates 16-bit immediate offset, wasting 1 code byte:
06BA:0131 098A4600          OR      [BP+SI+0046],CX

CJam, 162 bytes
KFmfbcACZbbcDsXbcSFseesicHsXbcEsXbcSFseesicAsXbcIsXbcDsXbcDFmfbcSHAmfbcBsXbceuHseesicGseesicSGseesicHsXbcAsXbcGseesicHAmfbcFseesicSHsXbcIsXbcFseesicSciBbIbcHAmfbc

Try it online!
Each letter is outputted individually and all the source codes are concatenated. The individual codes are:
  (1 byte)
S   e# Space

' (6 bytes)
HAmfbc
HA       e# Push 17 and 10    17 10
  mf     e# Factorization     17 [2 5]
    b    e# Base conversion   39
     c   e# Convert to char   ''

, (6 bytes)
DFmfbc
DF       e# Push 13 and 15    13 15
  mf     e# Factorization     13 [3 5]
    b    e# Base conversion   44
     c   e# Convert to char   ',

. (8 bytes)
SciBbIbc
Sci        e# Push space as an integer   32
   Bb      e# Convert to base 11         [2 10]
     Ib    e# Convert from base 18       44
       c   e# Convert to char            '.

A (6 bytes)
KFmfbc
KF       e# Push 20 and 15    20 15
  mf     e# Factorization     20 [3 5]
    b    e# Base conversion   65
     c   e# Convert to char   'A

B (7 bytes)
BsXbceu
Bs        e# Push 11 as a string   "11"   (i.e. [49 49])
  Xb      e# Convert from base 1   98
    c     e# Convert to char       'b
     eu   e# Uppercase             'B

a (5 bytes)
AsXbc
As      e# Push 10 as a string   "10"   (i.e. [49 48])
  Xb    e# Convert from base 1   97
    c   e# Convert to char       'a

The codes for d, e, h and i work exactly the same way as for a. The codes are:
DsXbc -> 'd
EsXbc -> 'e
HsXbc -> 'h
IsXbc -> 'i

n (6 bytes)
ACZbbc
ACZ      e# Push 10, 12 and 3   10 12 3
   b     e# Base conversion     10 [1 1 0]
    b    e# Base conversion     110
     c   e# Convert to char     'n

s (7 bytes)
Fseesic
Fs        e# Push 15 as a string   "15"
  ee      e# Enumerate             [[0 '1] [1 '5]]
    s     e# Convert to string     "0115"
     ic   e# Convert to char       's

The codes for t and u do the same thing but staring with G(16) and H(17):
Gseesic -> 't
Hseesic -> 'u

Most of these codes were found using another Cjam program to brute force all possible outputs for a given formula. If you want to use it, type some formula such as RTsbc, and it will show you how to get to each possible output by replacing R, T and U in the input by some 1 byte constant.


text:
Find longest consecutive characters in 2D array
Challenge
Given a 2D array, find the length and direction of all the longest consecutive characters.

If there are ties, output every possible winning combination in any order, (winners can overlap with other winners)
Zero "0" is a special padding character which cannot be a winner.
Code golf rules apply, shortest code wins.

Output: n,m,d
(Where n is the identified character of length m in d direction)
Valid characters for d: /|\-. See below examples to interpret meaning.
Input: The 2D array input should be in either of the following format, (your preference):
Option 1:
123
456
789

Option 2*:
[["123"],["456"],["789"]]

*or any combination of "{[()]}" characters can be used, with whitespace characters ignored outside of quotes. Quotes can be single or double. The array must read left-to-right, top-to-bottom.
Assumptions

The shape of the array will always be rectangular or square
There will always be at least one non-zero character
Array characters will only be alphanumeric

Examples
1.
0000000
0111100
0000000

1,4,-

2.
00000000
aaaa0000
00bbbbb0

b,5,-

3.
0y00x000
aayxaaaz
00xy000z
0x00y00z
x0000y0z

x,5,/
y,5,\

*(in any order)
4.
1111
1111
1111

1,4,-
1,4,-
1,4,-

5.
ccc
cwc
ccc

c,3,-
c,3,-
c,3,|
c,3,|

*(in any order)
6.
9

9,1,.

7.
ab
cd

a,1,.
b,1,.
c,1,.
d,1,.

*(in any order)

Sorry, I don't quite understand the .. What's the expected result of a 2x2 square of 3 1s and a 0?

@Neil The dot just means it doesn't have a direction because the longest string of characters, is of length one. A 2x2 matrix of 3 1s and a 0, would be: "1,2,| and 1,2,-"

Are we restricted to those 4 characters for d or can we use any 4 characters of our choosing?

Why restrict the input format to a string, instead of just saying it is a 2D array and letting the site defaults take care of it?

Suggested test case: ab\ncd\nea → [[a,1,.],[b,1,.],[c,1,.],[d,1,.],[e,1,.],[a,1,.]] (contains two [a,1,.], which is a pretty annoying edge case).

Can we use an alternative to the character ‘0’ for our spacer? e.g. the number 0 (if a list of lists is acceptable input per @JoKing’s comment), or a space character?

Jelly, 52 54 bytes
,UŒD€;,Z$ṣ€€”0ŒrẎ€⁺;€"“\/-|”Ẏ2ị$ÐṀ¹F}ḟ”0;€1,”.ʋØ½œị’Ɗ?

Try it online!
A monadic link that takes a list of strings and returns. list of maximal outputs each in the order n, m, d. A third of the code is to deal with the requirement for dots (last example). Now handles Kevin Cruijssen’s edge case correctly.

Fails for edge-case ["ab","cd","ea"]. It only outputs a single ['a', 1, '.'] triplet instead of two. Nice answer regardless, so +1 from me.

@KevinCruijssen I had thought about that, though wasn’t entirely clear what the desired output was. Should be fixable.

@KevinCruijssen Fixed.

Nice, and a lot shorter than I thought. :) Just 2 bytes.

Charcoal, 110 bytes
≔⪫Ａ⸿θＰθＦθ«≔⟦⟧ηＦ∧¬№0⸿ι⟦→↘↓↙⟧«≔ＫＤＬθ✳κζＷ›Ｌζ№ζι≔⊟ζε⊞η⟦κιＬζ⟧»¿⁼¹⌈Ｅη§κ²⊞υ⊟ηＦη⊞υκι»⎚≔⌈Ｅυ§ι²εＦΦυ⁼⊟ιε«⊟ι,Ｉε,Ｐ✳⊟ι⎇⊖ε¹.¦⸿

Try it online! Link is to verbose version of code. Takes an array of strings as input. Explanation:
≔⪫Ａ⸿θＰθ

Join the strings with carriage returns (not newlines) and output them without moving the cursor.
Ｆθ«

Loop over each character in the string.
≔⟦⟧η

Collect all of the consecutive characters starting at this position in an array.
Ｆ∧¬№0⸿ι⟦→↘↓↙⟧«

If the current character is not a 0 or a carriage return, then check all four directions.
≔ＫＤＬθ✳κζ

Peek as much as possible in that direction.
Ｗ›Ｌζ№ζι≔⊟ζε

Shorten the string until all of the characters are the same.
⊞η⟦κιＬζ⟧»

Save the results in the array.
¿⁼¹⌈Ｅη§κ²⊞υ⊟ηＦη⊞υκ

If the maximum consecutive character was 1, then only push one of the results to the final array, otherwise push all of them.
ι»

Reprint the current character, this time allowing the cursor to move.
⎚

Clear the input from the canvas.
≔⌈Ｅυ§ι²ε

Find the longest consecutive length.
ＦΦυ⁼⊟ιε«

Loop over the entries with that length.
⊟ι,Ｉε,

Output the character and length.
Ｐ✳⊟ι⎇⊖ε¹.¦⸿

If the length is 1 then output a . otherwise output a direction character. Then move to the start of the next line for the next result.

05AB1E, 82 bytes
í‚εεygÅ0«NFÁ]€ø`IIø)εεγJ0K}˜D€gøNUε"/\-|"XèªćÙš]€`D€ÅsZQÏεDÅsi¤'.:]{γεÐнÅsig4÷∍]€`

Input as a matrix of characters. Outputs as a list of triples.
Try it online or verify all test cases.
Explanation:
Generate all diagonals and anti-diagonals of the input-matrix:
í                 # Reverse each row of the (implicit) input-matrix
 ‚                # Pair it with the (implicit) input-matrix
  ε               # Map both matrices to:
   ε              #  Map each row to:
    yg            #   Get the length of the current row
      Å0«         #   Append that many 0s at the end of the row
         NFÁ      #   Rotate the row the 0-based index amount of times towards the right
  ]               # Close the rotate-loop and nested maps
   €ø             # Transpose/zip (swapping rows/columns) for both matrices
     `            # And push both matrices to the stack

Put them in a list with the input-matrix itself, as well as all columns of the input-matrix:
I                 # Push the input-matrix
 Iø               # Push the input-matrix, and transpose/zip (swapping rows/columns)
   )              # Wrap everything on the stack into a list

Now that we have a list of all diagonals, all anti-diagonals, all rows, and all columns, we're going to extract chunks of the same characters, and transform them into the output triplets:
ε                 # Map each matrix to:
 ε                #  Map each row to:
  γ               #   Split into chunks of the same elements
   J              #   Join them together to a string
    0K            #   Remove all 0s
 }˜               #  After the map of the rows: deep-flatten all chunks
    €g            #  Get the length of each chunk
   D  ø           #  And pair it with the chunk itself
       NU         #  Store the current index in variable `X`
       ε          #  Map over the chunk and length pairs:
        "/\-|"Xè  #   Get the `X`'th character from the string "/\|-"
                ª #   And append it to the pair
        ć         #   Extract head; push the length plus character pair and chunk
                  #   separately to the stack
         Ù        #   Uniquify the chunk so a single character remains
          š       #   And prepend it back in front of the length & character pair
]                 # Close both maps
 €`               # And then flatten once

Now that we have all our triplets, we only leave the largest ones:
 €Ås              # Get the middle element (the chunk-length) of each
    Z             # Get the maximum chunk-length (without popping the list itself)
     Q            # Check for each chunk-length if it's equal to this maximum
D     Ï           # And only leave the triplets at the truthy indices

And finally we have to fix all . edge cases, for chunk-lengths of size 1:
ε                 # Map each triplet:
 DÅsi             #  If the middle element (the chunk-length) is exactly 1:
     ¤'.:        '#   Replace the trailing character of the triplet with a "."
]                 # Close the if-statement and map
 {                # Sort the triplets
  γ               # And group them by the exact same triplets
   ε              # Map each chunk of the same triplets to:
    Ð             #  Triplicate the chunk of triplets
     нÅsi         #  If the middle element of each triplet is 1:
         g        #   Get the amount of the same triplets in this chunk
          4÷      #   Integer-divide it by 4
            ∍     #   And shorten the chunk of triplets to that size
   ]              # Close the if-statement and map
    €`            # Flatten once
                  # (after which the result is output implicitly)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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