Hi,

I would like to found a regex match in a stdout

stdout

 /dev/loop0: [2081]:64 (/a/path/to/afile.dat)

I would like to match

/dev\/loop\d/

and return /dev/loop0

but the \d seem not working with awk … ?

How to achieve this ? ( awk is not mandatory )

  • TwilightKiddy@programming.devEnglish
    2·
    3 days ago

    Assuming you made a bit of a typo with your regexp, any of these should work as you want:

    grep -oE '/dev/loop[0-9]+'
awk 'match($0, /\/dev\/loop[0-9]+/) { print substr($0, RSTART, RLENGTH) }'
sed -r 's%.*(/dev/loop[0-9]+).*%\1%'

    AWK one is a bit cursed as you can see. Such ways of manipulating text is not exactly it’s strong suite.