How to remove first bash argument and pass the others to another command?

Question

In bash $@ contains all the arguments used to call the script but I am looking for a solution to remove the first one

./wrapper.sh foo bar baz ...:

 #!/bin/bash

 # call `cmd` with bar baz ... (withouyt foo one)

I just want to call cmd bar baz ...

Answer 1

You can use shift to shift the argument array. For instance, the following code:

#!/bin/bash
echo "$@"
shift
echo "$@"

produces, when called with 1 2 3 prints 1 2 3 and then 2 3:

$ ./example.sh 1 2 3
1 2 3
2 3