Normally when a parameter is passed to a shell script, the value goes into ${1} for the first parameter, ${2} for the second, etc.
How can I set the default values for these parameters, so that if no parameter is passed to the script, we can use a default value for ${1}?
You can't, but you can assign to a local variable like this: ${parameter:-word} or use the same construct in the place you need $1. this menas use word if _paramater is null or unset
Note, this works in bash, check your shell for the syntax of default values
${parameter-word} - It will not clobber empty strings.#!/bin/sh
MY_PARAM=${1:-default}
echo $MY_PARAM
You could consider:
set -- "${1:-'default for 1'}" "${2:-'default 2'}" "${3:-'default 3'}"
The rest of the script can use $1, $2, $3 without further checking.
Note: this does not work well if you can have an indeterminate list of files at the end of your arguments; it works well when you can have only zero to three arguments.
Perhaps I don't understand your question well, yet I would feel inclined to solve the problem in a less sophisticated manner:
! [[ ${1} ]] && declare $1="DEFAULT"
Hope that helps.
$1 is not set, the name in declare $1="DEFAULT" is empty, which is not going to help. If it is set (for example, the script was invoked as ./script abc), then the declare you suggest makes the variable abc equal to DEFAULT without touching $1. You simply cannot set 'positional parameters' ($1, $2, etc) that way.