Process all arguments except the first one (in a bash script)

Question

I have a simple script where the first argument is reserved for the filename, and all other optional arguments should be passed to other parts of the script.

Using Google I found this wiki, but it provided a literal example:

echo "${@: -1}" 

I can't get anything else to work, like:

echo "${@:2}" 

or

echo "${@:2,1}" 

I get "Bad substitution" from the terminal.

What is the problem, and how can I process all but the first argument passed to a bash script?

Answer 1

Use this:

echo "${@:2}" 

---

The following syntax:

echo "${*:2}" 

would work as well, but is not recommended, because as @Gordon already explained, that using *, it runs all of the arguments together as a single argument with spaces, while @ preserves the breaks between them (even if some of the arguments themselves contain spaces). It doesn't make the difference with echo, but it matters for many other commands.

Answer 2

If you want a solution that also works in /bin/sh try

first_arg="$1" shift echo First argument: "$first_arg" echo Remaining arguments: "$@" 

shift [n] shifts the positional parameters n times. A shift sets the value of $1 to the value of $2, the value of $2 to the value of $3, and so on, decreasing the value of $# by one.