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?

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.

You can also use cut:
bash has some nice string handling; try this for instance:

great bebo! This is really 'Zen or the art of shell programming'! always appreciate short and simple but spectacular solutions! Keep it up!

bruce

Thanks a lot guys.

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

$(ls /vmware/number.* | awk 'BEGIN { FS = "." } ; { print $2 }' )

Spectacular? What?! ;):D shell-jutsu, but still not at dan grade :p:)