How to remove first bash argument and pass the others to another command?
bash, parameter-passing
Solution
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
Problem
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 ...`