bash - passing optional arguments to script - parametername + string_value

Question

i use a script that accepts parameter. parameters are optional and may occur in any order.

#!/bin/bash
# script name: test.sh
for var in "$@"
do
    if [ ! -z "$var" ]   && ([ $var = "--example" ] || [ $var = "-e" ]); then
        echo "example"
    elif [ ! -z "$var" ]   && ([ $var = "--project" ] || [ $var = "-p" ]); then
        echo "project with string xxxxxxx"
    fi
done

in this simple example, you could call it like follows (some examples):

# this will echo example
./test.sh --example 
# this will echo project with string xxxxxxx
./test.sh --project
# this will echo both example and project with string xxxxxxx
./test.sh --example --project

NOW, what i want to achieve is that i can do something like this (warning, this is pseuco code):

#!/bin/bash
# script name: test.sh
for var in "$@"
do
    if [ ! -z "$var" ]   && ([ $var = "--example" ] || [ $var = "-e" ]); then
        echo "example"
    elif [ ! -z "$var" ]   && ([ $var = "--project" ] || [ $var = "-p" ]); then
        echo "project with string $VAR_VALUE"
    fi
done



# this will echo example
./test.sh --example 
# this will echo project with string myproject1
./test.sh --project="myproject1" 
# this will echo both example and project with string myproject2
./test.sh --example --project="myproject2"

can someone help me rewrite it so this will work somehow?

John Kugelman · Accepted Answer · 2019-11-13 12:49:41Z

2

Use getopt. It handles short and long options, allows for both --long value and --long=value, decomposes -abc into -a -b -c, understands -- to end option parsing, and more.

#!/bin/bash

args=$(getopt -o ep: -l example,project: -n "$0" -- "$@") || exit
eval set -- "$args"

while [[ $1 != '--' ]]; do
    case "$1" in
        -e|--example) echo "example";      shift 1;;
        -p|--project) echo "project = $2"; shift 2;;

        # shouldn't happen unless we're missing a case
        *) echo "unhandled option: $1" >&2; exit 1;;
    esac
done
shift  # skip '--'

echo "remaining non-option arguments: $@"

edited Nov 13, 2019 at 12:49

answered Nov 13, 2019 at 12:34

John Kugelman

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2 Comments

jmrah Over a year ago

What does the getopt -n option do? I looked up the man page, but don't understand it.

John Kugelman Over a year ago

@jrahhali If the user makes a mistake with the arguments getopt will print "progname: error message". -n tells it what "progname" is so it looks like the error message is coming from your program.

dash-o · Accepted Answer · 2019-11-13 12:52:50Z

There are two possible path toward parsing argument list

Build custom option parser
use getopt, using 'long options'

The first approach is relatively simple (at this time). Using case instead of if to handle variants:

last_arg=
for arg in "$@"
do
    if [ "$last_arg" = "-p" ] ; then
        VAR_VALUE=$arg ;
        last_arg=
        echo "project with string $VAR_VALUE"
        continue
    fi
    case "$arg" in
        -e | --example)
            echo "example" ;;
        -p)
            last_arg=$arg ;;
        --project=*)
            VAR_VALUE=${arg#*=}
            echo "project with string $VAR_VALUE" ;;
        *) ERROR-MESSAGE ;;
    esac
done
exit

The BETTER approach is to leverage existing code. In particular getopt, which can handle long options:

#! /bin/bash
if T=$(getopt -o ep: --long 'example,project:' -n ${0#*/} -- "$@") ; then
        eval set -- "$T"
else
        exit $?
fi
while [ "$#" -gt 0 ] ; do
        case "$1" in
                -e | --example)
                        echo "example"
                        ;;
                -p | --project)
                        shift
                        VAR_VALUE=$1
                        echo "project with string $VAR_VALUE"
                        ;;
                --)
                        break
                        ;;
                *) echo "ERROR:$1" ;;
        esac
        shift
done

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