Hello @offshore9521 and All,
Humm…, @offshore9521, there are two separate problems with your regex !
From your INPUT text :
‘Colt?’ He turns to face her, and clears his throat. Guilty sign. Very guilty. Probably has a guilty look on his face, but you can’t tell, because he’s still wearing that helmet so you can’t see his eyes.Your regex (?<![\.\!\?])\r\n([a-z]) means :
If a \r\n sequence, followed with a letter (a through z), is not preceded by :
A dot
An exclamation mark
An interrogation mark
Then rewrite this letter
BTW, you can shorter this regex as (?<![.!?])\r\n([a-z])
But, the ’ character is not a character of the [.!?] list. Thus, this explains why your regex merges the lines 1 and 2
Now, I suppose that your true goal was : if a line is not a true sentence, then merge that line with the next one with a space character ? This lead to the following regex S/R :
FIND (?<![.!?’])\r\n(?=\w)
REPLACE \x20
So, any \r\n sequence, which respects the before and after conditions, is simply replaced with a space character
Notes :
As you see, I included the ’ character within the list of the forbidden chars, in the negative look-behind
I used \w, which is identical to the [\u\l\d_] class character, instead of [a-z] to not bother about case !
Remark :
Instead of \r\n, I could have used the \R syntax with matches any kind of line endings ( \r, \n or \r\n ), but, because it is preceded by a negative look-behind, we must insert \r and \n as forbidden characters, as well !FIND (?<![.!?’\r\n])\R(?=\w)
REPLACE \x20
Best Regards,
guy038
P.S. : As an exercice, try to understand why the following regex S/R does not work as expected :
FIND (?<![.!?’])\R(?=\w)
REPLACE \x20
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