Shell script

Check if at least n arguments are present

$# stores the number of arguments that were given when executing a shell script. Thus, it's possible to check if at least a given number of arguments are present.

The following fragment checks if at least 3 arguments were passed:

if [ $# -lt 3 ]; then
    echo Error: At least 3 arguments are required.
    exit 1
fi

# … rest of the shell script …

getopts

while getopts fx:h OPTION; do

  case $OPTION in
    f)
       echo f was specfied
       ;;
    x)
       echo x was specified with value $OPTARG
       ;;
    h)
       echo $0 [-f] [-x value] [-h]
       ;;
    esac

done

echo OPTIND=$OPTIND

$ ./script
OPTIND=1

$ ./script -h
./script [-f] [-x value] [-h]
OPTIND=2

$ ./script -x
./script: option requires an argument -- x
OPTIND=2

$ ./script -x foo
x was specified with value foo
OPTIND=3

$ ./script -xbar
x was specified with value bar
OPTIND=2

$ ./script -f -x baz
f was specfied
x was specified with value baz
OPTIND=4

The value of OPTIND can be used to «shift away» the option arguments so that the required arguments can be processed:

shift $((OPTIND-1))

Determine directory where script is located

dir="$(dirname "$(readlink -f "$0")")"

Check if program/command/executable is present

if command -v cmake &> /dev/null; then
    echo cmake was found
else
    echo cmake was not found
fi

Shell script

Check if at least n arguments are present

getopts

Determine directory where script is located

Check if program/command/executable is present

See also