Processing optional arguments with getopts in bash

Question

I have a bash script which processes an input file with optional arguments. The script looks like this

#!/bin/bash while getopts a:b:i: option do case "${option}" in a) arg1=${OPTARG};; b) arg2=${OPTARG};; i) file=${OPTARG};; esac done [ -z "$file" ] && { echo "No input file specified" ; exit; } carry out some stuff 

The script runs fine, but I need to specify the input file like so

sh script.sh -a arg1 -b arg2 -i filename 

I would prefer to be able to call the script without the -i option, like so

sh script.sh -a arg1 -b arg2 filename 

while still having the error message when no input file is specified. Is there a way to do this?

Answer 1

#!/bin/sh - # Beware variables can be inherited from the environment. So # it's important to start with a clean slate if you're going to # dereference variables while not being guaranteed that they'll # be assigned to: unset -v file arg1 arg2 # no need to initialise OPTIND here as it's the first and only # use of getopts in this script and sh should already guarantee it's # initialised. while getopts a:b:i: option do case "${option}" in (a) arg1=${OPTARG};; (b) arg2=${OPTARG};; (i) file=${OPTARG};; (*) exit 1;; esac done shift "$((OPTIND - 1))" # now "$@" contains the rest of the arguments if [ -z "${file+set}" ]; then if [ "$#" -eq 0 ]; then echo >&2 "No input file specified" exit 1 else file=$1 # first non-option argument shift fi fi if [ "$#" -gt 0 ]; then echo There are more arguments: printf ' - "%s"\n' "$@" fi 

I changed the bash to sh as there's nothing bash-specific in that code.

Answer 2

I would like to expand on Stéphane Chazelas' excellent answer in order to use bashism more on the question style, namely [ -z "$file" ] && { echo "No input file specified" ; exit; } -- instead of using ifs.

#!/bin/bash unset -v file arg1 arg2 while getopts "a:b:i:" option; do case "$option" in a ) arg1="$OPTARG";; b ) arg2="$OPTARG";; i ) file="$OPTARG";; * ) exit 1;; esac done shift "$((OPTIND - 1))" [ -z "$file" ] && [ "$#" -eq 0 ] && { echo "No input file" >&2; exit 1; } [ -z "$file" ] && [ "$#" -gt 0 ] && { file="$1"; shift; } echo "DEBUG: arg 1 is $arg1" echo "DEBUG: arg 2 is $arg2" echo "DEBUG: file is $file" [ "$#" -gt 0 ] && { echo "More arguments"; printf " - %s\n" "$@"; }