bash getopts not recognizing the second argument

Question

This is my entire script in its simplest form.

#!/bin/bash
src=""
targ=${PWD}

while getopts "s:t:" opt; do
  case $opt in
    s)
      src=$OPTARG
      ;;
    t)
      targ=$OPTARG
      ;;
  esac
  shift $((OPTIND-1))
done

echo "Source: $src"
echo "Target: $targ"

I run this script as getopts_test -s a -t b

However, it always prints the pwd in front of the Target: and never b

What am I missing here?

probably this might help http://stackoverflow.com/questions/16483119/example-of-how-to-use-getopts-in-bash — Naruto, Dec 05 '15 at 04:31
try taking the `shift` out of the while loop: move it to after `done` — glenn jackman, Dec 05 '15 at 04:36
When you do the shift inside the loop, you remove the `-s` and `a` arguments from the argument list, but `getopts` still thinks it processed arguments 1 and 2 (which now contain `-t` and `b`), so it looks beyond them and says "nothing else to do". — Jonathan Leffler, Dec 05 '15 at 04:44

score 5 · Accepted Answer · answered Dec 05 '15 at 04:44

The reason for why b is never printed is that the shift within the loop moves the processed options away after the first iteration, i.e. after a has been printed. Use of shift $((OPTIND-1)) is intended to access the possible given variadic parameters. Naturally, once you remove shift, targ gets reassigned to b, and ${PWD} is no longer included in it since you don't have concatenation of the strings (targ and the option of -t) anywhere.

sjsam · Answer 2 · 2015-12-05T06:23:39.900

3

An alternative to what @glenn-jackman suggested in his comment would be this :

#!/bin/bash
src=""
targ=${PWD}

while getopts "s:t:" opt; do
  case $opt in
    s)
      src=$OPTARG
      echo "Source: $src"
      ;;
    t)
      targ=$OPTARG
      echo "Target: $targ"
      ;;
  esac

done

shift $((OPTIND-1)) # Turning to non-option arguments say a file name and so on.

Here you go with the natural flow of arguments without shifting.

edited Dec 05 '15 at 06:23

answered Dec 05 '15 at 04:52

sjsam

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The shift is needed after the loop so that you can remove the option arguments (and `--`) and leave `"$@"` with just the non-option arguments. – Jonathan Leffler Dec 05 '15 at 06:01
@JonathanLeffler : Thankyou for reminding about the non-option args. Edited the answer. :) – sjsam Dec 05 '15 at 06:25

bash getopts not recognizing the second argument

2 Answers2