getting positional arguments in bash

Question

I have a bash script called foo with variable number of arguments, with the first one being a required one, i.e.:

foo a1 b2 b3 b4 ...

I understand that in bash $1 will get me argument a1, but is there a way to get all the rest of the arguments? $@ or $* seem to get me all the arguments including a1.

Answer 1

./foo.sh 1 2 3 4

#!/bin/bash
echo $1;
echo $2;
echo $3;
echo $4;

Will output:

1
2
3
4

Answer 2

Slice the $@ array.

echo "${@:2}"

Answer 3

#!/bin/sh

echo $*
shift
echo $*

shift will shift all the parameters, running the previous example would give:

$ test_shift.sh a b c d e
a b c d e
b c d e

Answer 4

You can use shift command for that. That will remove $1 and you can access the rest of arguments starting with $1.