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separating filename and extension

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# 1  
Old 04-06-2007
separating filename and extension

Hi (warning: newbie question),
I am writing a script to run a series of tests on a program, which involves a line:

Code:
for file in `ls test_suite/*.args`

but later I want to send the output to file.out. But I need to separate the filename and extension somehow...Also $file contains "test_suite/" before each entry as well, how do I get rid of that (I assume it's in a similar way as the separation above)

Thanks!
# 2  
Old 04-06-2007
HI,
You can get the filename alone excluding the directory using following command
for file in `ls test_suite/*.args`
do
F_NAME=$(basename ${file})
done

Thanks
Raghu
The Following User Says Thank You to Raghuram.P For This Useful Post:
mohit_iitk (02-13-2012)
# 3  
Old 04-06-2007
Code:
for f in test_suite/*.args;do 
	fn="${f%.*}" 
	printf "filename without path and extension: %s, extension only: %s\n" "${fn##*/}" "${f##*.}"  
done

With bash you can do something like this (if I understand correctly the requirement):
Code:
set test_suite/*.args
set "${@##*/}"
printf "%s\n" "${@%.*}">file.out


Last edited by radoulov; 04-06-2007 at 07:40 AM..
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