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regexlinuxgreplvm

grep regex how to remove single quote

I'd like to get the quoted string from output of lvscan which is :

  ACTIVE            '/dev/mysys/root' [297.46 GiB] inherit

What I've done is to use code below:

lvscan | grep -o "[^']\+\b'"

and what I got is :

/dev/mysys/root'

if I remove the very last single quote, I got three lines:

ACTIVE
/dev/mysys/root
[297.46 GiB] inherit

I also tried:

lvscan | grep -o "[@(^'\+.*)].*\>'"
lvscan | grep -o "[^']\+\>'"

What did I miss here?

Any comments would appreciate. Thank you.

Solution

  • if you have -P with grep you can use

    lvscan | grep -oP "(?<=')[^']+(?=')"