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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'
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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/'
Thor
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3I had to add the
-ror--regexp-extendedoption otherwise I was gettinginvalid reference \1 ons' 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
-rto the above you should get an error saying "invalid reference". What version of sed are you using? – Thor Aug 11 '14 at 20:46 -
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1How does this answer the question of "only the string [that matched]". Matching the whole line was not in the question. – Jan 29 '18 at 15:45
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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 '18 at 15:43
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@Oliver: That is because of the
-rwhich interprets the RE as an ERE. The default is to use BRE. See this comparison for more detail – Thor Feb 13 '19 at 12:43 -
echo "atetb" | sed 's/.*\(test\).*/\1/'returns a non match, because it is substitution and not match – Kjeld Flarup Nov 12 '20 at 08:55 -
sed?grep's-oswitch looks like a shorter and cleaner way:echo "atestb" | grep -o 'test'. – manatwork Jul 16 '12 at 07:54echo "aSomethingYouWantb" | sed -En 's/a(.*)b/\1/p'– luckman212 Oct 17 '20 at 00:59