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 )

  • pelya@lemmy.world
    4·
    3 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
      01·
      3 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?