Shell Programming and Scripting

Hi,

sample datas are :

drwxr-xr-x 2 beewin abidev 96 Jun 13 2006 bwin
drwxrwxr-x 2 blsmg01 smg 96 Jun 13 2006 blsmg01
drwxr-xr-x 2 ccmdummy ccm_root 8192 Jun 13 2006 ccmdum
drwxr-xr-x 5 dipayan users 8192 Oct 29 09:05 dip

I want to cut the last field.

I use ls -ll | grep "dipayan" | cut -d" " -f22 to cut dip from
drwxr-xr-x 5 dipayan users 8192 Oct 29 09:05 dip.

but f22 is not working for other cases as the no of spaces are different for different data.
Is there any cut option so that always
ls -ll | grep "dipayan" | cut -d" " -f22 command cut the last field.
i.e. cut bwin for beewin and like that.

It looks like there are 9 fields, separated by spaces, not 22. Do you want to display only the last field?
Then use: cut -d" " -f9

Thanks for reply.
I already tried with that but it returns "users".

Problem with the data is, there are multiple spaces between fields and varys from data to data.
Is there any way to convert multiple spaces between fields to single space.

Ah! Try this:

Code:

dir ... | sed 's/.* //'

That should delete everything up to and including the last space on the line.

in place of the cut, try awk

awk '{print $9}'

this will print the 9th field, which was separated by space and/or tab characters.

Thanks a lot, it's working fine.