Don't need the whole line, just the match from regular expression

Question

I simply need to get the match from a regular expression:

$ cat myfile.txt | SOMETHING_HERE "/(\w).+/"

The output has to be only what was matched, inside the parenthesis.

Don't think I can use grep because it matches the whole line.

Please let me know how to do this.

Answer 1

Use the -o option in grep.

Eg:

$ echo "foobarbaz" | grep -o 'b[aeiou]r'
bar

Answer 2

If you want only what is in the parenthesis, you need something that supports capturing sub matches (Named or Numbered Capturing Groups). I don't think grep or egrep can do this, perl and sed can. For example, with perl:

If a file called foo has a line in that is as follows:

/adsdds      /

And you do:

perl -nle 'print $1 if /\/(\w).+\//' foo

The letter a is returned. That might be not what you want though. If you tell us what you are trying to match, you might get better help. $1 is whatever was captured in the first set of parenthesis. $2 would be the second set etc.

Answer 3

This will accomplish what you are requesting, but I don't think it is what you really want. I put the .* in the front of the regex to eat up anything before the match, but that is a greedy operation, so this only matches the penultimate \w character in the string.

Note that you need to escape the parens and the +.

sed 's/.*\(\w\).\+/\1/' myfile.txt