How to find all words appearing between parenthesis?

Question

I have a file containing some words in parenthesis. I'd like to compile a list of all of the unique words appearing there, e.g.:

This is some (text).
This (text) has some (words) in parenthesis.
Sometimes, there are numbers, such as (123) in parenthesis too.

This would be the resulting list:

text
words
123

How can I list all of the items appearing between parenthesis?

Answer 1

You can use awk like this:

awk -F "[()]" '{ for (i=2; i<NF; i+=2) print $i }' file.txt

prints:

text
text
words
123

You can use an array to print the unique values:

awk -F "[()]" '{ for (i=2; i<NF; i+=2) array[$1]=$i; print array[$1] }' file.txt

prints:

text
words
123

HTH

Answer 2

With GNU grep, you can use a perl-compatible regex with look-around assertions to exclude the parens:

grep -Po '(?<=\().*?(?=\))' file.txt | sort -u

Answer 3

grep -oE '\([[:alnum:]]*?\)' | sed 's/[()]//g' | sort | uniq

• -o Only prints the matching text
• -E means use extended regular expressions
• \( means match a literal paren
• [[:alnum:]] is the POSIX character class for letters and numbers.

That sed script should strip out the parens. This is tested against GNU grep, but BSD sed so be wary.