Shell Programming and Scripting

Hi,
I am working on a parsing script but cannot figure out how to accomplish this. Here is a simplified version of the script:

Code:

#!/bin/bash
DS=$1
DS=`expr $DS \* 2`
DS=`expr $DS + 7`

cat $FILENAME | awk '/<row><v> +[0-9]/' | awk '{printf("%.0f %.0f\n", $6, $9)}'

The problem is that I want the $9 (the 9th column) to instead be the "DS"-th column. For example, if DS=2, then it should be $11. Any ideas on how to specify the variable name ($9) to be determined by another variable?

I would really appreciate any advice.
Thanks.
- Nisrak

Try this:

Code:

awk -vDS=$DS '/<row><v> +[0-9]/ { printf("%.0f %.0f\n", $6, $(9+DS))}' $FILENAME

That did the trick! Thanks for the help.
- Nisrak