CODEWITHSUNDEEP
bash
Kiran Mohan
Kiran Mohan

Reputation: 3006

bash script: error in conditional statement

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

Upvotes: 0

Views: 37

Answers (2)

PM 2Ring
PM 2Ring

Reputation: 55469

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.

Upvotes: 2

Dan
Dan

Reputation: 101

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

Upvotes: 0

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