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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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5 Answers 5

Use the -o option in grep.

Eg:

$ echo "foobarbaz" | grep -o 'b[aeiou]r'
bar
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1  
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
6  
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
1  
However, that is a very good thing to know anyways :-) –  Kyle Brandt Aug 6 '09 at 18:00
1  
@KyleBrandt: To match only one part (e.g.: the parenses) it's possible to mark the rest with a look ahead or look behind: (?<= ) and (?= ) –  DrYak Jan 20 at 13:09

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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2 Things:

  • As stated by @Rory, you need the -o option, so only the match are printed (instead of whole line)
  • In addition, you neet the -P option, to use Perl regular expressions, which include useful elements like Look ahead (?= ) and Look behind (?<= ), those look for parts, but don't actually match and print them.

If you want only the part inside the parensis to be matched:

grep -oP '(?<=\/\()\w(?=\).+\/)' myfile.txt

if file contains the sting /(a)5667/, grep will print 'a', because:

  • /( are found by \/\(, but because they are in a look-behind (?<= ) they are not reported
  • a is matched by \w and is thus printed (because of -o )
  • )5667/ are found b< \).+\/, but because they are in a look-ahead (?= ) they are not reported
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Because you tagged your question as bash in addition to shell, there is another solution beside grep :

Bash has its own regular expression engine since version 3.0, using the =~ operator, just like Perl.

now, given the following code:

#!/bin/bash
DATA="test <Lane>8</Lane>"

if [[ "$DATA" =~ \<Lane\>([[:digit:]]+)\<\/Lane\> ]]; then
        echo $BASH_REMATCH
        echo ${BASH_REMATCH[1]}
fi
  • Note that you have to invoke it as bashand not just sh in order to get all extensions
  • $BASH_REMATCH will give the whole string as matched by the whole regular expression, so <Lane>8</Lane>
  • ${BASH_REMATCH[1]} will give the part matched by the 1st group, thus only 8
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