15
regex and awk...
(programming.dev)
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 -E for '+' 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 00002018
And 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: 00000027
And 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.
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.
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.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].
this post was submitted on 08 Jun 2025
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