How to print help if no argument passed in bash [duplicate]

Question

I have created a script in order to use arguments while running that script. How to check if there were no arguments provided ? It must shows print help if no argument passes.

while test -n "$1"; do case "$1" in -help|-h) print_help exit $ST_UK ;; --version|-v) print_version $PROGNAME $VERSION exit $ST_UK ;; --activeusers|-a) opt_var=$2 au shift;; --dailyusers|-d) opt_var1=$2 dau shift;; *) echo "Unknown argument: $1" print_help exit $ST_UK ;; esac shift done

Use: (( $# )) || print_help the the top of your script

anubhava
– anubhava

2018-08-27 10:59:31 +00:00
Commented Aug 27, 2018 at 10:59 — anubhava
– anubhava, Commented Aug 27, 2018 at 10:59
Great! It's working

blaCkninJa
– blaCkninJa

2018-08-27 11:09:11 +00:00
Commented Aug 27, 2018 at 11:09 — blaCkninJa
– blaCkninJa, Commented Aug 27, 2018 at 11:09

Toby Speight · Accepted Answer · 2018-08-27 11:05:22Z

5

You can do it the same way you would for any POSIX shell, by testing the $# (number of arguments) magic variable:

if [ "$#" -eq 0 ] then usage >&2 exit 1 fi

answered Aug 27, 2018 at 11:05

Toby Speight

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2 Comments

glenn jackman Over a year ago

Or, if no arguments, set the args to "--help" and use the while loop: [[ $# -eq 0 ]] && set -- -h

Toby Speight Over a year ago

@glenn - less good, because you want the error case to write to stderr and exit with a failure, but you want requested help to write to stdout and exit success. That's why I prefer a usage function for this.

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