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 )
grep -o '/dev/loop[0-9]\+'Why are you making a literal out of the + operator? This will not work.
grep -o ‘/dev/loop[0-9]+’Because it does work, you need
grep -Efor ‘+’ to work without escaping. Also, your quotes are wrong, ‘ should be ’ .
Also, your quotes are wrong, ‘ should be ’ .
Alright, I am getting confused. What quotes are those?
I got this from the stuff I copied from your comment:
❯ ./UTF8txt2hex ’‘ UTF-8: e2 80 99 e2 80 98 UTF-16: 2019 2018 UCS 4: 00002019 00002018And these are the single quote and backtick I used (of course I had to escape them, because they are the actual ones):
❯ ./UTF8txt2hex \` UTF-8: 60 UTF-16: 60 UCS 4: 00000060 ❯ ./UTF8txt2hex \' UTF-8: 27 UTF-16: 27 UCS 4: 00000027And from what I see, your original comment had the correct ones, so was this all to elicit this response out of me?
You could try [0-9] instead?
awk '/\/dev\/loop[0-9]/ {print}'If you have a larger sample of input and desired output, people can help you better.
I know this isn’t grep or awk but of you simply want the first part I would probably use cut as following: ``` cut -d : -f 1
Simply put, cut the line in multiple parts with the colon as the delimiter and choose the first part.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.
