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regex and awk... (programming.dev)

submitted 4 weeks ago by [email protected] to c/[email protected]

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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] 1 points 4 weeks ago

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.