How to skip the first argument in $@?

Question

My code:

#!/bin/bash  for i in $@;     do echo $i; done; 

run script:

# ./script 1 2 3  1 2 3 

So, I want to skip the first argument and get:

# ./script 1 2 3  2 3 

Answer 1

Use the offset parameter expansion

#!/bin/bash  for i in "${@:2}"; do     echo $i done 

Example

$ func(){ for i in "${@:2}"; do echo "$i"; done;}; func one two three two three 

Answer 2

Use shift command:

FIRST_ARG="$1" shift REST_ARGS="$@"