Shell Programming and Scripting

Hi,

I am using ls command as below to get the list of *.edi files from 2 different subdirectories as1 & as2:

/ntekupal/test$ ls ./as1/*.edi ./as2/*.edi

It is giving me list of edi files in these folders as below :

./as1/6df667.edi ./as2/823ssd3.edi

Now my requirement is to just get the list of files without showing the absolute path like below :

6df667.edi 823ssd3.edi

Thanks
Raj.

see if you systems supports basename

then you could pipe this to xargs basename to get what you want.

I am new to unix so not sure about usage of xargs.
Can you please elaborate on that usage.

Thanks.

Code:

ls ./as1/*.edi ./as2/*.edi | xargs basename

Code:

printf "%s\n" ./as1/*.edi ./as2/*.edi | awk -F/ '{print $NF}'

Or:

Code:

cd as1 && printf '%s\n' *.edi
cd as2 && printf '%s\n' *.edi

Using xargs the way PORTER stated above will not work for both the files.

$ ls ./as1/*.edi ./as2/*.edi
./as1/jks.edi ./as2/rs.edi

$ ls ./as1/*.edi ./as2/*.edi | xargs basename
jks.edi


So this is the way :

$ ls -l ./as1/*.edi ./as2/*.edi | awk -F '/' '{print $NF'}
jks.edi
rs.edi

or

$ ls ./as1/*.edi ./as2/*.edi | while read file; do basename $file; done
jks.edi
rs.edi

What's the point of using the -l option? You only want the filenames.

For that matter, why use ls at all?