Bash – How to write a very simple wrapper that provides default parameters

bashparametershell-script

Given a program that requires some parameters, e.g. program -in file.in -out file.out, what would be the simple-most approach to write a bash script that could be called with or without any of these parameters and use default values for each?

script -in otherfile would run program -in otherfile -out file.out,
script -out otherout -furtherswitch would run program -in file.in -out otherout -furtherswitch etc.

Best Answer

A default value is easy to define in Bash:

foo="${bar-default}" # Sets foo to the value of $bar if defined, "default" otherwise
foo="${bar:-default}" # Sets foo to the value of $bar if defined or empty, "default" otherwise

To process your parameters, you can use a simple loop:

while true
do
    case "${1-}" in
        -in)
            infile="${2-}"
            shift 2
            ;;
        -out)
            outfile="${2-}"
            shift 2
            ;;
        *)
            break
            ;;
    esac
done

program -in "${infile-otherfile}" -out "${outfile-otherout}" "$@"

Useful reads:

I also recommend using getopt instead, because it is able to handle many special cases which would very quickly complicate and clutter your code (Non-trivial example).

Related Question