grep: regex only for match anything between parenthesis

Question

How do I match word only between parenthesis

Input : this is (test.com)

Desire Output : test.com

Answer 1

Here are a few options, all of which print the desired output:

1. Using grep with the -o flag (only print matching part of line) and Perl compatible regular expressions (-P) that can do lookarounds:

   printf "this is (test.com)\n" | grep -Po '(?<=\().*(?=\))'
   

   That regex might need some explaining:

   • (?<=\() : This is a positive lookbehind, the general format is (?<=foo)bar and that will match all cases of bar found right after foo. In this case, we are looking for an opening parenthesis, so we use \( to escape it.

   • (?=\)) : This is a positive lookahead and simply matches the closing parenthesis.

2. The -o option to grep causes it to only print the matched part of any line, so we look for whatever is in parentheses and then delete them with sed:

   printf "this is (test.com)\n" | grep -o '(.*)' | sed 's/[()]//g'
   

3. Parse the whole thing with Perl:

   printf "this is (test.com)\n" | perl -pe 's/.*\((.+?)\)/$1/'
   

4. Parse the whole thing with sed:

   printf "this is (test.com)\n" | sed 's/.*(\(.*\))/\1/'
   

Answer 2

One approach would be to use PCRE - Perl Compatible Regular Expressions with grep:

$ echo "this is (test.com)" | grep -oP '(?<=\().*(?=\))'
test.com

References

• regular-expressions.info - lookarounds