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Extracting a regex matched with 'sed' without printing the surrounding characters

How do I make this only print test:

echo "atestb" | sed -n 's/\(test\)/\1/p'
  • 7
    Must use sed? grep's -o switch looks like a shorter and cleaner way: echo "atestb" | grep -o 'test'.
    – manatwork
    Jul 16, 2012 at 7:54
  • In case you are trying to output only the matching portion BETWEEN two known strings, try echo "aSomethingYouWantb" | sed -En 's/a(.*)b/\1/p'
    – luckman212
    Oct 17, 2020 at 0:59

1 Answer 1


You need to match the whole line:

echo "atestb" | sed -n 's/.*\(test\).*/\1/p'


echo "atestb" | sed 's/.*\(test\).*/\1/'
  • 3
    I had to add the -r or ` --regexp-extended` option otherwise I was getting invalid reference \1 on s' command's RHS ` error. Aug 11, 2014 at 16:12
  • @danielsokolowski: this example uses basic regular expressions (BRE) and thus capture groups are noted with escaped parentheses. If you add -r to the above you should get an error saying "invalid reference". What version of sed are you using?
    – Thor
    Aug 11, 2014 at 20:46
  • 1
    How does this answer the question of "only the string [that matched]". Matching the whole line was not in the question.
    – user56041
    Jan 29, 2018 at 15:45
  • 3
    @jww: Did you test the answer? You need to match the whole line because it is a substitution command, i.e. you substitute the whole line with what was matched
    – Thor
    Jan 30, 2018 at 15:43
  • 1
    I had to use sed -rn 's/.*(MYPATTERN).*/\1/p'
    – Oliver
    Feb 12, 2019 at 21:03

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