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 )
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.
You’ve got lots of answers, so I’ll just say that shorthand character classes like \s, \w, and \d - all those backslash ones - are not widely supported, especially in the only POSIX tools. Many tools have an extended or Perl mode that makes them available, but some don’t. You can’t rely on them being everywhere. That’s why you’re getting suggestions to use explicit, long-form character classes like [0-9].
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.
