bash - How to replace the nth column/field in a comma-separated string using sed/awk?

Question

assume I have a string

"1,2,3,4"

Now I want to replace, e.g. the 3rd field of the string by some different value.

"1,2,NEW,4"

I managed to do this with the following command:

echo "1,2,3,4" | awk -F, -v OFS=, '{$3="NEW"; print }'

Now the index for the column to be replaced should be passed as a variable. So in this case

index=3

How can I pass this to awk? Because this won't work:

echo "1,2,3,4" | awk -F, -v OFS=, '{$index="NEW"; print }'
echo "1,2,3,4" | awk -F, -v OFS=, '{$($index)="NEW"; print }'
echo "1,2,3,4" | awk -F, -v OFS=, '{$$index="NEW"; print }'

Thanks for your help!

Answer 1

Have the shell interpolate the index in the awk program:

echo "1,2,3,4" | awk -F, -v OFS=, '{$'$index'="NEW"; print }'

Note how the originally single quoted awk program is split in three parts, a single quoted beginning '{$', the interpolated index value, followed by the single quoted remainder of the program.