20
regex and awk... (programming.dev)
submitted 1 week ago by [email protected] to c/[email protected]

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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[-] [email protected] 6 points 1 week ago

Regex syntax and features vary between implementations. \d isn't supported by BRE/ERE flavors.

GNU grep supports PCRE, so you can use grep -oP '/dev/loop\d' or grep -o '/dev/loop[0-9]' if you are matching only one digit character.

[-] [email protected] 4 points 1 week ago

I wish there was one single unifying regex standard.

(obligatory xkcd in 3..2..)

[-] [email protected] 6 points 1 week ago

Use [0-9] to match the number

[-] [email protected] 5 points 1 week ago

Or, alternatively, [[:digit:]], and dont' forget to add a quntifier + to match multiple digits. See documentaion for details.

awk '/^\/dev\/loop[[:digit:]]+/{print}'
[-] [email protected] 4 points 1 week 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

[-] [email protected] 0 points 1 week ago* (last edited 1 week ago)

That would also print the colon

Edit: missed the separator token. Sorry guys

[-] [email protected] 3 points 1 week ago

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

[-] [email protected] 3 points 1 week ago

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

[-] [email protected] 2 points 1 week ago

Why would it print the colon?

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

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