Array index from first to second last in PowerShell

Question

How can I get the array element range of first to second last?

For example,

$array = 1,2,3,4,5
$array[0] - will give me the first (1)
$array[-2] - will give me the second last (4)
$array[0..2] - will give me first to third (1,2,3)
$array[0..-2] - I'm expecting to get first to second last (1,2,3,4) but I get 1,5,4 ???

I know I can do long hand and go for($x=0;$x -lt $array.count;$x++), but I was looking for the square bracket shortcut.

Answer 1

You just need to calculate the end index, like so:

$array[0..($array.length - 2)]

Do remember to check that you actually have more than two entries in your array first, otherwise you'll find yourself getting duplicates in the result.

An example of such a duplicate would be:

@(1)[0..-1]

Which, from an array of a single 1 gives the following output

1 
1

Answer 2

As mentioned earlier the best solution here: $array[0..($array.length - 2)]

The problem you met with $array[0..-2] can be explained with the nature of "0..-2" expression and the range operator ".." in PowerShell. If you try to evaluate just this part "0..-2" in PowerShell you will see that result will be an array of numbers from 0 to -2.

>> 0..-2
0
-1
-2

And when you're trying to do $array[0..-2] in PowerShell it's the same as if you would do $array[0,-1,-2]. That's why you get results as 1, 5, 4 instead of 1, 2, 3, 4.

It could be kind of counterintuitive at first especially if you have some Python or Ruby background, but you need to take it into account when using PowerShell.

Answer 3

There might be a situation where you are processing a list, but you don't know the length. Select-object has a -skiplast parameter.

(1,2,3,4,5 | select -skiplast 2)

1
2
3