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My input:

   1:FAILED      +  *1      0     (8328832,AR,UNDECLARED)

This is what I expect:

8328832,AR,UNDECLARED

I am trying to find a general regular expression that allows to take any content between two brackets out.

My attempt is

  grep -o '[(.*?)]' test.txt > output.txt

but it doesn't match anything.

See Question&Answers more detail:os

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Still using and

grep -oP '(K[^)]+' file

K means that use look around regex advanced feature. More precisely, it's a positive look-behind assertion, you can do it like this too :

grep -oP '(?<=()[^)]+' file

if you lack the -P option, you can do this with :

perl -lne '/(K[^)]+/ and print $&' file

Another simpler approach using

awk -F'[()]' '{print $2}' file

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