Pass command line parameters to a program inside the shell script

Question

As for ./script.sh arg1 [arg2 arg3 ...], the command line arguments arg1, arg2, ... can be got by $1, $2, ... But the number of arguments is NOT fixed.

In the shell script, I want to pass the arguments starting from arg2 to a program,

#/bin/bash
...
/path/to/a/program [I want to pass arg2 arg3 ... to the program]
...

How could I do it since there could be one or more arguments?

Solution 1

The usual way would be to save a copy of arg1 ("$1") and shift the parameters by one, so you can refer to the whole list as "$@":

#!/bin/sh
arg1="$1"
shift 1
/path/to/a/program "$@"

bash has some array support of course, but it is not needed for the question as posed.

If even arg1 is optional, you would check for it like this:

if [ $# != 0 ]
then
    arg1="$1"
    shift 1
fi

Solution 2

You can slice the positional parameters using parameter expansion. The syntax is:

${parameter:offset:length}

If length is omitted it is taken as till the last value.

As you were to pass from second to last arguments, you need:

${@:2}

Example:

$ foo() { echo "${@:2}" ;}

$ foo bar spam egg
spam egg

Comments

[Marius]

The double-quotes with $@ tells the shell to double-quote each parameter (something nice to do if the parameters contain interesting characters such as parentheses or asterisks). If you don't care about that, a plain $* works...

[Gilles 'SO- stop being evil']

@Lee Yes, the double quotes are necessary. Otherwise, instead of passing through the list of arguments, each argument is split at whitespace, then each piece is interpreted as a wildcard pattern and, if the pattern matches, it's replaced by the list of matches. Generally, speaking, always double quote variable substitutions.