1:FAILED + *1 0 (8328832,AR,UNDECLARED)
This is what I expect : 8328832,AR,UNDECLARED
I am trying to find a general regex expression that allows to take any content between two brackets out. My attempt is.
grep -o '\[(.*?)\]' test.txt > output.txt
But it cannot match anything =(
Thanks
Still using grep and regex
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 :
perl -lne '/\(\K[^\)]+/ and print $&' file
Another simpler approach using awk
awk -F'[()]' '{print $2}' file
