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in reply to Re: Regex to add space after punctuation sign
in thread Regex to add space after punctuation sign

Thank you, Abigail. Can you please explain how does the replacement part of your expression work? If I understand correctly, /e evaluates $1 || "$2 " -- that way if $1 is not empty, it should not add second part?

--dda

P.S. I added end-of-line processings as was suggested by aragorn: 
s/(\d[,.]\d)|([^\w\s](?!\s)(?!$))/$1 || "$2 "/ge;

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Re: Regex to add space after punctuation sign
by Abigail-II (Bishop) on Jan 08, 2004 at 09:58 UTC
    Evaluating $1 || "$2 " means you get $1 if $1 is true (which means, anytime \d[.,]\d matches), otherwise you get $2 (the punctation character), followed by a space.

    I would handle end-of-line processing as follows: 

    s/(\d[,.]\d)|([^\w\s])(?=\S)/$1 || "$2 "/ge;

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    A single positive look-ahead is, IMO, more clear than two negative look-aheads.

    Abigail

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      Thank you. You are the best!

      --dda

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