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Possible Duplicate:
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'
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    Must use sed? grep's -o switch looks like a shorter and cleaner way: echo "atestb" | grep -o 'test'. – manatwork Jul 16 '12 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 '20 at 0:59
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You need to match the whole line:

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

or

echo "atestb" | sed 's/.*\(test\).*/\1/'
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    I had to add the -r or ` --regexp-extended` option otherwise I was getting invalid reference \1 on s' command's RHS ` error. – Daniel Sokolowski Aug 11 '14 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 '14 at 20:46
  • sed (GNU sed) 4.2.2 on cygwin – Daniel Sokolowski Aug 12 '14 at 4:42
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    How does this answer the question of "only the string [that matched]". Matching the whole line was not in the question. – user56041 Jan 29 '18 at 15:45
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    @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 '18 at 15:43

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