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Is there a way to change the command line arguments in a Bash script? For example, a Bash script is invoked like this:
./foo arg1 arg2 Is there a way to change the value of arg1 within the script? Something like:
$1="chintz" 
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$1 means an input argument and -z means non-defined or empty. You're testing whether an input argument to the script was defined when running the script. Follow this answer to receive notifications.
You can handle command-line arguments in a bash script in two ways. One is by using argument variables, and another is by using the getopts function.
Passing Arguments Using Flags and Options in Shell ScriptAn option is always followed by a value while flags are not followed by any value. First, we will make a new bash script that takes two different arguments (options) i.e. -n/--name for name, and -i/--id for an identification number.
You have to reset all arguments. To change e.g. $3:
$ set -- "${@:1:2}" "new" "${@:4}" Basically you set all arguments to their current values, except for the one(s) that you want to change. set -- is also specified by POSIX 7.
The "${@:1:2}" notation is expanded to the two (hence the 2 in the notation) positional arguments starting from offset 1 (i.e. $1). It is a shorthand for "$1" "$2" in this case, but it is much more useful when you want to replace e.g. "${17}".
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