Shell Script Argument without value

Question

I'm creating a very simple sh file to do something, but I want to pass an argument without a value, for example:

./fuim -l

But I receive the following message:

./fuim: option requires an argument -- l

If I pass a random value, like ./fuim -l 1, it works perfectly. How can I do that?

Here's what I have so far:

while getopts e:f:l:h OPT
do
    case "$OPT" in
        h) print_help ;;
        e) EXT=$OPTARG ;;
        f) PROJECT_FOLDER=$OPTARG ;;
        l) LIST_FILES=1 ;;
        ?) print_help ;;
    esac
done

shift $((OPTIND-1))

if [ -z "$EXT" ] || [ -z "$PROJECT_FOLDER" ]; then
    print_help
fi

So it's a mandatory argument option. Optional argument options use two colons (:) instead of one (as in getopts e:f:l::h), look the manual of getopts(1) for that, as there's some side effects when you use optional argument options. — Luis Colorado
– Luis Colorado, Commented Aug 13, 2014 at 6:37

Barmar · Accepted Answer · 2014-08-12 18:13:25Z

51

When using getopts, putting : after an option character means that it requires an argument. So change l: to l, as in:

while getopts e:f:lh OPT

This makes it a boolean option, either it's there or it isn't.

answered Aug 12, 2014 at 18:13

Barmar

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2 Comments

Aman Jain Over a year ago

Went through multiple blog, but no one has mentioned about the : behaviour. Thanks for explaining.

Barmar Over a year ago

Those must be pretty bad tutorials, as this is one of the most basic features.

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