bash script: error in conditional statement

Question

What is wrong with the below script?

if [ $# -eq 0 -o $1 = "-h" -o $1 = "--help" ]; then    ## line no: 11
    echo $usage
    exit 0;
fi

If run without any arguments, it gives the below error

% ./test.sh
./test.sh: line 11: [: too many arguments

You should quote your variables, particularly $1 in this case. May I recommend shellcheck.net ? — Biffen
– Biffen, Commented Feb 2, 2015 at 7:14

PM 2Ring · Accepted Answer · 2015-02-02 09:26:13Z

2

Rather than using -o you can use || to do short-circuit evaluation, i.e., if a condition is true the subsequent conditions won't be evaluated.

if [ -z "$1" ] || [ "$1" = '-h' ] || [ "$1" = '--help' ]; then
    echo "$usage"
    exit 0
fi

If you're only running this script in bash you can do use this more compact syntax:

if [[ -z "$1" || "$1" = '-h' || "$1" = '--help' ]]; then
    echo "$usage"
    exit 0
fi

but that may not work in other shells.

Note that I've also quoted the argument to echo. In general, you should quote parameters unless you explicitly want word splitting. I guess I ought to mention that printf is more robust than echo, but echo is ok for fixed strings.

edited Feb 2, 2015 at 9:26

answered Feb 2, 2015 at 8:10

PM 2Ring

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Comments

Dan · Accepted Answer · 2015-02-02 07:26:12Z

0

You have to quote $1 because if it's empty and not quoted, it will be replaced by literally nothing instead of and empty string

if [ $# -eq 0 -o "$1" = "-h" -o "$1" = "--help" ]; then
    echo $usage
    exit 0;
fi

answered Feb 2, 2015 at 7:26

Dan

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