Question

I have created simple script:

#!/bin/sh
column=${1:-1}
awk '{colawk='$column'+1; print colawk}'

But when I run:

ls -la | ./Column.sh 4

I receive output:
5
5
5
5

But I have expected receive 5th column. Why this error?

Answer 1

I believe this will do what you’ve attempted in your example:

#!/bin/sh
let "column=${1:-1} + 1"
awk "{print \$$column}"

However, I don’t see why you’re adding one to the column index? You’ll then not be able to intuitively access the first column.

I’d to it this way instead:

#!/bin/sh
let "column=${1:-1}"
awk "{print \$$column}"

The argument to ./Column.sh will be the column number you want, 0 will give you all columns, while a call without arguments will default the column index to 1.

---

I know bash. I would like make arithmetic with AWK

In that case, how about:

#!/bin/sh
column=${1:-1}
awk 'BEGIN{colawk='$column'+1} {print $colawk}'

Or, simply:

#!/bin/sh
awk 'BEGIN{colawk='${1:-1}'+1} {print $colawk}'

Two things I changed in your script:

1. put the arithmetic in a BEGIN{} block since it only needs to be done once and not repeated for every input line.
2. “print $colawk” instead of “print colawk” so we’re printing the column indexed by colawk instead of its value.