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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.

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3 Answers 3

Use the -o option in grep.


$ echo "foobarbaz" | grep -o 'b[aeiou]r'
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Good grief... Do you have any idea how many times I wrestled with sed backreferences to do that? –  Insyte Aug 6 '09 at 17:36
The o option to grep/egrep returns only what matched the entire regular expression, not just what is in () like he asked for. –  Kyle Brandt Aug 6 '09 at 17:59
However, that is a very good thing to know anyways :-) –  Kyle Brandt Aug 6 '09 at 18:00

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.

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I was just trying to match what is in parenthesis. Seems like passing it to a perl or a php script might be the answer. –  Alex L Aug 6 '09 at 18:01

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
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