Only return the matched string in sed [duplicate]

Question

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'

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

Thor · Accepted Answer · 2012-07-15 21:41:13Z

73

You need to match the whole line:

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

or

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

answered Jul 15, 2012 at 21:41

Thor

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3

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, 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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