extract all words between two match words

Question

I Have a single line as follows, I need all the words between select and Done

 vertical on; select blah blah blah contains all special characters including /*?&;  Done

Answer 1

Using bash regular expression would give (assuming the line in a variable):

$ line="vertical on; select blah blah blah contains all special characters including /*?&;  Done"
$ [[ "$line" =~ select(.*)Done ]] && echo ${BASH_REMATCH[1]}
blah blah blah contains all special characters including /*?&;

Answer 2

Try this,

 sed -e 's/.*select//;s/Done.*//'

Answer 3

Try

sed -e 's/select\(.*\)Done/\1/'

Answer 4

Using Perl-compatible regular expressions (PCRE):

(?<=word1).*?(?=word2)

Applying the above pattern with GNU grep:

$ grep -Po '(?<=select).*?(?=Done)' <<< ' vertical on; select blah blah blah contains all special characters including /*?&;  Done'
 blah blah blah contains all special characters including /*?&;  

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Explanation

• (?<=word1): Match word1 before the current position but don't include it in the result.
• .*?: Match any string.
• (?=word2): Match word2 after the current position but don't include it in the result.

The (?<=word1) and (?=word2) patterns are collectively known as lookaround. This feature is not supported by POSIX engines (BRE and ERE) so you need a more powerful engine (such as PCRE).