1mo ago

regex and awk...

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

Hi,
I would like to found a regex match in a stdout
stdout
 text
    
 /dev/loop0: [2081]:64 (/a/path/to/afile.dat)

  
I would like to match
 regex
    
/dev\/loop\d/


  
and return /dev/loop0
but the \d seem not working with awk ... ?
How to achieve this ? ( awk is not mandatory )

9 comments

Use [0-9] to match the number
- Or, alternatively, [[:digit:]], and dont' forget to add a quntifier + to match multiple digits. See documentaion for details.
  
  awk '/^\/dev\/loop[[:digit:]]+/{print}'
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.
- I wish there was one single unifying regex standard.
  (obligatory xkcd in 3..2..)
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
- That would also print the colon
  Edit: missed the separator token. Sorry guys
  
  The field separator is declared to be the colon, with -F:, so the fields end and start at colons.
  
  No, because we're telling to use : as a separator with the -F flag
  
  Why would it print the colon?

9 comments