How to get the nth positional argument in bash?

Question

How to get the nth positional argument in bash?

Thanks.

Edit: I forgot to say but I meant that n is a variable.

Answer 1

$1 $2 ... $n

$0 contains the name of the script.

Answer 2

$n       

Answer 3

As you can see in the Bash by Example, you just need to use the automatic variables $1, $2, and so on.

$# is used to get the number of arguments.

Answer 4

Read

Handling positional parameters

and

Parameter expansion

$0: the first positional parameter

$1 ... $9: the argument list elements from 1 to 9

Answer 5

If N is saved in a variable, use

eval echo \${$N}

if it's a constant use

echo ${12}

since

echo $12

does not mean the same!

Answer 6

Use Bash's indirection feature:

#!/bin/bash
n=3
echo ${!n}

Running that file:

$ ./ind apple banana cantaloupe dates

Produces:

cantaloupe

Edit:

You can also do array slicing:

echo ${@:$n:1}

but not array subscripts:

echo ${@[n]}  #  WON'T WORK