如何跳过$ @中的第一个参数？

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

Answer 3

Look into Parameter Expansions in the bash manpage.

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

Answer 4

You could just have a variable testing whether it's the first argument with something like this (untested):

#!/bin/bash
FIRST=1
for i in $@
do
    if [ FIRST -eq 1 ]
    then
        FIRST=0
    else
        echo $i
    fi
done