20
regex and awk... (programming.dev)
submitted 4 days ago by Corsair@programming.dev to c/linux@lemmy.ml

cross-posted from: https://programming.dev/post/31833654

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 )

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[-] mmmm@sopuli.xyz 5 points 4 days ago

Not sure if I'm understanding, but can't you just pipe the whole thing to awk and capture the first field? Like

echo "/dev/loop0: [2081]:64 (/a/path/to/afile.dat)" | awk -F: '{print $1}'

Which would print

/dev/loop0

[-] lungdart@lemmy.ca 0 points 4 days ago* (last edited 2 days ago)

That would also print the colon

Edit: missed the separator token. Sorry guys

[-] mmmm@sopuli.xyz 3 points 3 days ago

No, because we're telling to use : as a separator with the -F flag

[-] manxu@piefed.social 3 points 4 days ago

The field separator is declared to be the colon, with -F:, so the fields end and start at colons.

[-] learnbyexample@programming.dev 2 points 4 days ago

Why would it print the colon?

this post was submitted on 08 Jun 2025
20 points (95.5% liked)

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