bash - 如何跳过 $@ 中的第一个参数？

Question

我的代码:

#!/bin/bash

for i in $@;
    do echo $i;
done;

运行脚本:

# ./script 1 2 3

1
2
3

所以，我想跳过第一个参数并得到:

# ./script 1 2 3

2
3

Answer 1

使用offset参数展开

#!/bin/bash

for i in "${@:2}"; do
    echo $i
done

例子

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