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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T00:14:15+00:00 2026-05-14T00:14:15+00:00

Hi I need a regular expression that’ll give me something like this part ./something\,

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Hi I need a regular expression that’ll give me something like this part

./something\", [something.sh

from something like this string

("./something\", [something.sh", ["./something\", [something.sh"], [/* 37 vars */])

is that possible? I’m having real trouble making this since there’s that \” escape sequence and also that ‘,’ character, so I cannot simply use match everything instead of these characters.
I’m working on unix so it’s also possible to use pipeline of few greps or something like that.
Thanks for advice.

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  1. Editorial Team
    Editorial Team
    2026-05-14T00:14:15+00:00Added an answer on May 14, 2026 at 12:14 am

    With Perl you can use Text::Balanced which has an extract_quotelike function to do what you need.

    You can do it manually with:

     /"((?:\\.|.)*?)"/
    

    Basically: starting from a quote, if you see a \, grab the next character (even if it’s a quote), else continue until you see a quote.

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