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Old 07-27-2004, 02:57 PM   #1
mehesque
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I need to parse a word: awk or sed?


I just need a one-liner. I search this directory for a file named finished.[then a number] "finished.3". It will never go over 9. I just want to grab the number from the end of the file name to put it in an if statement.

$(ls /dir/number.* | %%%%% )

I'm thinking either awk or sed, but I've never run across anything that parses words. It can be based on the the characters of the line too...

Thoughts?
 
Old 07-27-2004, 03:16 PM   #2
druuna
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Grepping the number-part from a string (3 if finished.3)

Using sed:
ls finished.[0-9] | sed 's/finished\.\([0-9]\)/\1/'

Using awk:
ls finished.[0-9] | awk -F"." '{ print $2 }'

There are probably many more ways to do this.

Hope this helps.
 
Old 07-27-2004, 03:26 PM   #3
Bebo
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You can also use cut:
Code:
ls finished.[0-9] | rev | cut -c1
bash has some nice string handling; try this for instance:
Code:
for FILE in `ls finished.[0-9]` ; do echo ${FILE##*\.} ; done
 
Old 07-27-2004, 03:36 PM   #4
bruce ford
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Quote:
ls finished.[0-9] | rev | cut -c1
great bebo! This is really 'Zen or the art of shell programming'! always appreciate short and simple but spectacular solutions! Keep it up!

bruce
 
Old 07-27-2004, 03:48 PM   #5
mehesque
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Thanks a lot guys.

I ended up going with awk and changing the field separator.

$(ls /vmware/number.* | awk 'BEGIN { FS = "." } ; { print $2 }' )
 
Old 07-27-2004, 04:23 PM   #6
Bebo
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Spectacular? What?! shell-jutsu, but still not at dan grade
 
  


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