regex - How to extract the content between two brackets by using grep? -

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   1:failed      +  *1      0     (8328832,ar,undeclared)

this expect : 8328832,ar,undeclared

i trying find general regex expression allows take content between 2 brackets out. attempt is.

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

but cannot match =(

thanks

still using , 

grep -op '\(\k[^\)]+' file

\k means use look around regex advanced feature. more precisely, it's positive look-behind assertion, can :

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

if lack -p option, can :

perl -lne '/\(\k[^\)]+/ , print $&' file

another simpler approach using 

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

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