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Old 10-11-2007, 02:27 AM   #1
ratul_11
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Cut command Problem


Hi All
I am using ll(long listing) and the output produced by the ll i want to cut the last field by using cut command. But as u know that there is not any separator
and the spaces between each column are not fixed. How to solve it.

regards
Anirban Adhikary.
 
Old 10-11-2007, 02:42 AM   #2
colucix
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Use awk instead, as in
Code:
ls -l | awk '{print $NF}'
this will print the last field of each line, despite the number of blank spaces between fields. However it does not work if filenames contain blank spaces. Please note that the output of the command above is exactly the same as
Code:
ls -1
with no problems for filenames!
 
Old 10-11-2007, 02:44 AM   #3
druuna
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Hi,

I'm not sure if you want the last field printed or remove it from the rest of the line.

colucix shows how to print the last field.

If you want all but the last field:

sed 's/\(.*\) .*$/\1/' infile

Hope this helps.
 
Old 10-11-2007, 03:02 AM   #4
jschiwal
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Another way to do it is to compress the extra spaces and use a space as the delimiter:

Code:
 ls -l | tr -s ' ' | cut -d' ' -f 8
I'm not saying this is better than the awk method presented earlier.

However, why do you want to use a long listing and only keep the name. Might as well use a short listing "ls -1" or use find, "find ./ -maxdepth 1 -type f"
 
  


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