Handling unused getopts argument

Question

I have a script that starts with getopts and looks as follows:

USAGE() { echo -e "Usage: bash $0 [-w <in-dir>] [-o <out-dir>] [-c <template1>] [-t <template2>] \n" 1>&2; exit 1; }

if (($# == 0))
then
    USAGE
fi

while getopts ":w:o:c:t:h" opt
do
    case $opt in
        w ) BIGWIGS=$OPTARG
        ;;
        o ) OUTDIR=$OPTARG
        ;;
        c ) CONTAINER=$OPTARG
        ;;
        t ) TRACK=$OPTARG
        ;;
        h ) USAGE
        ;;
        \? ) echo "Invalid option: -$OPTARG exiting" >&2
        exit
        ;;
        : ) echo "Option -$OPTARG requires an argument" >&2
        exit
        ;;
    esac
done

more commands etc

echo $OUTDIR
echo $CONTAINER

I am fairly new to getopts. I was doing some testing on this script and at some stage, I didn't need/want to use the -c argument [-c ]. In other words, I was trying to test another specific part of the script not involving the $CONTAINER variable at all. Therefore, I simply added # in front of all commands with the $CONTAINER and did some testing which was fine.

When testing the script without using $CONTAINER, I typed:

bash script.bash -w mydir -o myoutdir -t mywantedtemplate

However, I was wondering, given my getopts command I didn't get a warning. In other words, why did I not get a warning asking for -c argument. Is this possible? Does the warning only occur if I type:

bash script.bash -w mydir -o myoutdir -t mywantedtemplate -c

UPDATE

After doing some testing, I think that is it:

If you don't explicitly write "-c", getopts won't "ask" you for it and give you an error (unless your script is doing something with it - i.e. if you haven't put # in front of each command using this argument)
You only get an error if you put "-c "

Is this correct?

score 2 · Accepted Answer · answered Sep 14 '18 at 19:42

getopts does not warn when some options are not used (i.e. they are optional). Usually that's a good thing because some options (e.g. -h) are not used with other options. There is no way to specify mandatory options directly with the Bash builtin getopts. If you want mandatory options then you will need to write code to check that they have been used. See bash getopts with multiple and mandatory options. Also (as you have found), you won't get an error if you fail to write code to handle options specified in the optstring (first) argument to getopts.

You could get a kind of automatic warning for mandatory arguments by using the nounset setting in your Bash code (with set -o nounset or set -u). That would cause warnings to be issued for code like echo $CONTAINER if the -c option is not specified so $CONTAINER is not set. However, using the nounset option would mean that all of your code needs to be written more carefully. See How can I make bash treat undefined variables as errors?, including the comments and "Linked" answers, for more information.

score 1 · Answer 2 · answered Nov 24 '21 at 11:02

You might use this script:

printf captures the stdout and put into the stderr and then back to the stdout also captures the exit code, so we can handle if the code is greater than 0.

some_command() {
    echo 'this is the stdout'
    echo 'this is the stderr' >&2
    exit 1
}

run_command() {
    {
        IFS=$'\n' read -r -d '' stderr;
        IFS=$'\n' read -r -d '' stdout;
        (IFS=$'\n' read -r -d '' _ERRNO_; exit ${_ERRNO_});
    } < <((printf '\0%s\0%d\0' "$(some_command)" "${?}" 1>&2) 2>&1)
}

echo 'Run command:'
if ! run_command; then
    ## Show the values
    typeset -p stdout stderr
else
    typeset -p stdout stderr
fi

Just replace some_command with getopts ":w:o:c:t:h"

Handling unused getopts argument

2 Answers2