I need to run a script to remove the last column of different comma separated files.
The problem is that the number of columns of my files will be different and I won't know that number every time i run my script.
Is there any command I can use to remove the last column without specifying its number?
I can use cut but only when I know the column position.
I was trying another approach. If I calculate the number of the last column then I could use cut.
To find the last column I use:
but now I have the problem of how do I store that value in a variable so I can use cut?
If I use:
I get this error: bash: FS=,: command not found
Thanks in advance
loperam
Last edited by Scott; 09-03-2010 at 03:36 PM..
Reason: Please use code tags
bartus11
When using awk -F\, -vOFS=\, '{$NF=$NF-1}1' file
I get my file but the last column shows the value -1
What I want is that last column to be removed, not to show any other value
Scottn
I don't need to store all the last comma-separated files in one variable.
What I was trying to do as a second approach was to identify the position of that last column and store that single position number in a variable so I could use cut to remove it
What else can I use to get that position number?
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