When I type:

echo "foo bar" | gawk -v var="content" '{print $var}'

I expect the output to be content. Why is it foo bar?

  • 8
    Try removing $ from $var
    – taliezin
    Jun 21 '16 at 13:17
  • Indeed. I got confused with bash syntax...
    – nicoco
    Jun 21 '16 at 13:30
  • It's a good question and though the answer shows that the $ was misplaced I'm not sure it should be closed as typo. I've seen several of these recently so perhaps we need a canonical one
    – roaima
    Jul 19 '16 at 22:31
  • Didn't find this in time and created a duplicate question by accident.
    – Multisync
    Nov 3 '21 at 18:32

Variables are referenced by name as in var, not $var in awk. $n refers to the nth field: $1 for the first field, $2 for the second... or the whole record for n == 0 ($0 is the full record).

Those don't have to be literal numbers. You can use $(1+1) or $variable. If variable contains 1, then $variable will be the first field. A commonly used one is $NF for the last field (NF is the special variable that contains the number of fields).

If, like in your example, the variable doesn't contain a number, that non-number is understood as 0.

So var="content"; print $var, is the same as var=0; print $var, and thus the same as print $0, that is print the whole record.

What you want here is:

echo "foo bar" | gawk -v var="content" '{print var}'

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