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.

7 Answers 7


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 parenthesis to be matched, do the following:

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

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

  • /( are found by \/\(, but because they are in a look-behind (?<= ) they are not reported
  • a is matched by \w and is thus printed (because of -o )
  • )5667/ are found by \).+\/, but because they are in a look-ahead (?= ) they are not reported

Use the -o option in grep.


$ echo "foobarbaz" | grep -o 'b[aeiou]r'
  • 5
    Good grief... Do you have any idea how many times I wrestled with sed backreferences to do that?
    – Insyte
    Aug 6, 2009 at 17:36
  • 12
    The o option to grep/egrep returns only what matched the entire regular expression, not just what is in () like he asked for. Aug 6, 2009 at 17:59
  • 2
    However, that is a very good thing to know anyways :-) Aug 6, 2009 at 18:00
  • 2
    @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, 2015 at 13:09
  • this only works for me using egrep. Though it's been 12 years to probably a lot has changed. Jul 9, 2021 at 17:29
    sed -n "s/^.*\(captureThis\).*$/\1/p"

-n      don't print lines
s       substitute
^.*     matches anything before the captureThis 
\( \)   capture everything between and assign it to \1 
.*$     matches anything after the captureThis 
\1      replace everything with captureThis 
p       print it
  • This should be the accepted answer, IMHO. Apr 30, 2020 at 18:07

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:

DATA="test <Lane>8</Lane>"

if [[ "$DATA" =~ \<Lane\>([[:digit:]]+)\<\/Lane\> ]]; then
        echo $BASH_REMATCH
        echo ${BASH_REMATCH[1]}
  • 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
  • Dear @DrYak, I hope you're not parsing XML with regex here.. :)
    – joonas.fi
    Jun 16, 2016 at 12:43
  • It's even worse. I'm parsing a horrible mix of XML and FASTA data (which both use the > symbol for entirely different purposes) as spewed out by the SANSparallel fast largescale alignement software. Of course both formats are spewed interlaced without any escaping. So it's impossible to throw some standard XML library at this. And I'm using Bash regex at this point of the code because I only need to extract a couple of data, and 2 regex do the job much better for me than writing a dedicated parser for this mess. #LifeInBioinformatics
    – DrYak
    Jun 18, 2016 at 10:24
  • In other words: there's a point where extracting 1 single number is simpler to do with a regex rathan than dancing the whole XML tango
    – DrYak
    Jun 18, 2016 at 10:25
  • Hah, gotcha! :)
    – joonas.fi
    Jul 4, 2016 at 10:07
  • 1
    @Rajib: As far as I know, BASH only support POSIX RegEx but not PCRE . Thus there is a few features missing. Indeed, the "look ahead/look behinds" aren't available and using grouping is the next best thing.
    – DrYak
    Oct 30, 2020 at 19:10

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.

  • 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, 2009 at 18:01
  • Are you really sure sed can do named capturing groups ? I was unable to find anything related...
    – ssc
    May 11, 2020 at 16:01

Assuming the file contains:

$ cat file
Text-here>xyz</more text

And you want the character(s) between > and </ , you can use either:

grep -oP '.*\K(?<=>)\w+(?=<\/)' file
sed -nE 's:^.*>(\w+)</.*$:\1:p' file
awk '{print(gensub("^.*>(\\w+)</.*$","\\1","g"))}' file
perl -nle 'print $1 if />(\w+)<\//' file

All will print a string "xyz".

If you want to capture the digits of this line:

$ cat file

grep -oP '.*\K(?<=>)[0-9]+(?=<\/)' file
sed -E 's:^.*>([0-9]+)</.*$:\1:' file
awk '{print(gensub(".*>([0-9]+)</.*","\\1","g"))}' file
perl -nle 'print $1 if />([0-9]+)<\//' file

  • 1
    To me crucial was to realize \d doesn't work with sed. There's a reason you use [0-9]+ there. :)
    – user27432
    May 7, 2019 at 16:42
  • 1
    @user27423 It does not, but POSIX character classes (painful reading, pleasant reading) do: echo 'Text-<here>1234</text>-ends' | sed -E 's|.*>([[:digit:]]+)<.*|\1|'. In some cases (e.g. [0-9] vs. [[:digit:]]) they don't help legibility, in others I think they do (e.g. [ \t\n\r\f\v] vs. [:space:]). Jan 21, 2020 at 8:51
  • @SamuelHarmer Could you please clarify what is it that you mean with: It does not?
    – user427450
    Mar 11, 2020 at 15:52
  • @Isaac I was referring to @user27432's comment about the \d character group not working, and drawing their attention to POSIX character classes. Mar 11, 2020 at 18:17

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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