Extracting last parsed segment with the cut command

phudgens · #1 (**permalink**) 02-06-2009

I can see in the man pages how cut can be used to extract a specified string by using the -f qualifier. However, I can't find anywhere how to specify the "last" parsed segment with the cut command. If the string you are parsing is (or can be) differant each time the script is run, but you know that you always want the last segment parsed from the input string, how do you so specify in your command?

THanks
Phudgens
Denver

Gee-Money · #2 (**permalink**) 02-06-2009

instead of cut, you could use awk:

Code:

echo "one two three four" | awk '{print $NF}'

gives your "four", and:

Code:

echo "one two three" | awk '{print $NF}'

gives you "three".

The variable 'NF' by itself is the number of fields in the record (line), and if you prefix that with a '$' then that will give you the last field.

hth

phudgens · #3 (**permalink**) 02-09-2009

Thanks for the help. My final code ended up being:

set CurrentPwd = `pwd`
set CurrentFolder = `echo $CurrentPwd | awk -F'/' '{print $NF}'`

Paul H.

danmero · #4 (**permalink**) 02-09-2009

Quote:

Originally Posted by phudgens

Thanks for the help. My final code ended up being:

set CurrentPwd = `pwd`
set CurrentFolder = `echo $CurrentPwd | awk -F'/' '{print $NF}'`

Paul H.

What about parameter expansion:

Code:

set CurrentFolder = ${PWD##*/}

phudgens · #5 (**permalink**) 02-09-2009

Couldn't get that to work with either upper or lower case "pwd". Got a "Variable syntax" error message.

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