Using getopts inside a Bash function

Question

I'd like to use getopts inside a function that I have defined in my .bash_profile. The idea is I'd like to pass in some flags to this function to alter its behavior.

Here's the code:

function t() {     echo $*     getopts "a:" OPTION     echo $OPTION     echo $OPTARG } 

When I invoke it like this:

t -a bc 

I get this output:

-a bc ?   

What's wrong? I'd like to get the value bc without manually shifting and parsing. How do I use getopts correctly inside a function?

EDIT: corrected my code snippet to try $OPTARG, to no avail

EDIT #2: OK turns out the code is fine, my shell was somehow messed up. Opening a new window solved it. The arg value was indeed in $OPTARG.

Answer 1

As @Ansgar points out, the argument to your option is stored in ${OPTARG}, but this is not the only thing to watch out for when using getopts inside a function. You also need to make sure that ${OPTIND} is local to the function by either unsetting it or declaring it local, otherwise you will encounter unexpected behaviour when invoking the function multiple times.

t.sh:

#!/bin/bash  foo() {     foo_usage() { echo "foo: [-a <arg>]" 1>&2; exit; }      local OPTIND o a     while getopts ":a:" o; do         case "${o}" in             a)                 a="${OPTARG}"                 ;;             *)                 foo_usage                 ;;         esac     done     shift $((OPTIND-1))      echo "a: [${a}], non-option arguments: $*" }  foo foo -a bc bar quux foo -x 

Example run:

$ ./t.sh a: [], non-option arguments: a: [bc], non-option arguments: bar quux foo: [-a <arg>] 

If you comment out # local OPTIND, this is what you get instead:

$ ./t.sh a: [], non-option arguments: a: [bc], non-option arguments: bar quux a: [bc], non-option arguments: 

Other than that, its usage is the same as when used outside of a function.