Bash: parsing command line arguments with getopt

In this example we have a function that can optionally! accept the following parameters:

-v
-d
--debug
--name VALUE

The call to getoptdefines the accepted names. The-o dvdefines that we accept two short switches-vand-d. The --long debugdefines that we accept the--debugswitch. The--long name:defines that we accept the--name` option followed by some value.

Then we have a whole while loop to actually take out the values from the list on the comand line and to assign the appropriate values to variables. Either 1 to the variables representing the switches, or the actual value passed to the --name option.

At the end we need to call this function with the args $0 "$@" expression.

examples/shell/cli.sh

#!/bin/bash

function args()
{
    options=$(getopt -o dv --long debug --long name: -- "$@")
    if [ $? -ne 0 ]; then
        echo "Incorrect option provided"
        exit 1
    fi
    eval set -- "$options"
    while true; do
        case "$1" in
        -v)
            VERBOSE=1
            ;;
        -d)
            DEBUG=1
            ;;
        --debug)
            DEBUG=1
            ;;
        --name)
            shift; # The arg is next in position args
            NAME=$1
            ;;
        --)
            shift
            break
            ;;
        esac
        shift
    done
}

args $0 "$@"

echo $NAME
echo $DEBUG
echo $VERBOSE

An example calling it

$ ./cli.sh --name "Foo Bar" --debug -v

Foo Bar
1
1

$ ./cli.sh --wrong

getopt: unrecognized option '--wrong'
Incorrect option provided

If condition in bash

I used to have this, in the above code:

[ $? -eq 0 ] || {
    echo "Incorrect option provided"
    exit 1
}

This is I think the shell-style meaning:

"Either the previous command is successful (exit 0) or || do the block (echo and exit)".

Later I realized using if would make it much more readable so I changed it to:

if [ $? -ne 0 ]; then
    echo "Incorrect option provided"
    exit 1
fi

They both do the same.