Print out "string" between "match1" and "match2"

echo "string" | sed -e 's/.*match1//' -e 's/match2.*$//'
But who knows to delete the rest of the lines? I want only "string".

By: P17
2009-04-01 15:01:27

What Others Think

Not entirely sure what you're wanting, this maybe? sed -e 's/^.*\(string\).*$/\1/'
TheMightyBuzzard · 592 weeks and 2 days ago
This should be what you're looking for. Only prints lines with match1 and match2 in them, and only the section between the two matches. sed -ne 's/.*\(match1.*match2\).*/\1/p' or, a bit easier: egrep -o 'match1.*match2' Works with standard grep here, too, but YMMV.
Viaken · 592 weeks and 1 day ago
If you don't know what you're after, why are you posting to CLF? This isn't a support forum, but a repository of the best commands on the web. BTW: echo "string" | awk '/match1/,/match2/'
atoponce · 592 weeks and 1 day ago
I got it: sed -n '/match1/,/match2/p' file.txt | sed -e '/match1/d' | sed -e '/match2/d'
P17 · 589 weeks and 3 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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