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 )

  • @4am@lemm.ee
    1•6 days ago

    Why are you making a literal out of the + operator? This will not work.

    grep -o ‘/dev/loop[0-9]+’
    • pelya
      4•6 days ago

      Because it does work, you need grep -E for ‘+’ to work without escaping. Also, your quotes are wrong, ‘ should be ’ .

      • @ulterno@programming.devEnglish
        0•6 days ago

        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?