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

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?

http://wiki.bash-hackers.org/scripting/posparams

It explains the use of shift (if you want to discard the first N parameters) and then implementing Mass Usage

Came across this looking for something else. While the post looks fairly old, the easiest solution in bash is illustrated below (at least bash 4) using set -- "${@:#}" where # is the starting number of the array element we want to preserve forward:

    #!/bin/bash

    someVar="${1}"
    someOtherVar="${2}"
    set -- "${@:3}"
    input=${@}

    [[ "${input[*],,}" == *"someword"* ]] && someNewVar="trigger"

    echo -e "${someVar}\n${someOtherVar}\n${someNewVar}\n\n${@}"

Basically, the set -- "${@:3}" just pops off the first two elements in the array like perl's shift and preserves all remaining elements including the third. I suspect there's a way to pop off the last elements as well.

Working in bash 4 or higher version:

#!/bin/bash
echo "$0";         #"bash"
bash --version;    #"GNU bash, version 5.0.3(1)-release (x86_64-pc-linux-gnu)"

In function:

echo $@;              #"p1" "p2" "p3" "p4" "p5"
echo ${@: 0};  #"bash" "p1" "p2" "p3" "p4" "p5"
echo ${@: 1};         #"p1" "p2" "p3" "p4" "p5"
echo ${@: 2};              #"p2" "p3" "p4" "p5"
echo ${@: 2:1};            #"p2"
echo ${@: 2:2};            #"p2" "p3"
echo ${@: -2};                       #"p4" "p5"
echo ${@: -2:1};                     #"p4"

Notice the space between ':' and '-', otherwise it means different:

${var:-word} If var is null or unset,
word is substituted for var. The value of var does not change.

${var:+word} If var is set,
word is substituted for var. The value of var does not change.

Which is described in:Unix / Linux - Shell Substitution

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.