Hello guys

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:

cat filename | awk 'BEGIN {FS=","} END {print NF}'

position='cat filename | awk 'BEGIN {FS=","} END {print NF}' '

You can remove last column using just AWK:

awk -F\, -vOFS=\, '{NF=(NF>0)?NF-1:0}1' file

Hi.

Using NF will print the field number, not the value of the field. For the value, use $NF.

If I understand you correctly, you want to store all the last comma-separated fields in one variable?

$ position=$(sed "s/.*,//" file1)

Thanks for replying guys

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?

Thanks

Check my code. It is NF not $NF. Also use my updated version, so you don't have problems when file contains empty lines.

I apologise. I misread your question

that works as I needed!
thanks bartus11