How to convert string to integer in PowerShell

Use:

$filelist = @(11, 1, 2)
$filelist | sort @{expression={$_[0]}} | 
  % {$newName = [string]([int]$($_[0]) + 1)}
  New-Item $newName -ItemType Directory

Example:

2.032 MB (2,131,022 bytes)

$u=($mbox.TotalItemSize.value).tostring()

$u=$u.trimend(" bytes)") #yields 2.032 MB (2,131,022

$u=$u.Split("(") #yields `$u[1]` as 2,131,022

$uI=[int]$u[1]

The result is 2131022 in integer form.

You can specify the type of a variable before it to force its type. It's called (dynamic) casting (more information is here):

$string = "1654"
$integer = [int]$string

$string + 1
# Outputs 16541

$integer + 1
# Outputs 1655

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As an example, the following snippet adds, to each object in $fileList, an IntVal property with the integer value of the Name property, then sorts $fileList on this new property (the default is ascending), takes the last (highest IntVal) object's IntVal value, increments it and finally creates a folder named after it:

# For testing purposes
#$fileList = @([PSCustomObject]@{ Name = "11" }, [PSCustomObject]@{ Name = "2" }, [PSCustomObject]@{ Name = "1" })
# OR
#$fileList = New-Object -TypeName System.Collections.ArrayList
#$fileList.AddRange(@([PSCustomObject]@{ Name = "11" }, [PSCustomObject]@{ Name = "2" }, [PSCustomObject]@{ Name = "1" })) | Out-Null

$highest = $fileList |
    Select-Object *, @{ n = "IntVal"; e = { [int]($_.Name) } } |
    Sort-Object IntVal |
    Select-Object -Last 1

$newName = $highest.IntVal + 1

New-Item $newName -ItemType Directory

Sort-Object IntVal is not needed so you can remove it if you prefer.

[int]::MaxValue = 2147483647 so you need to use the [long] type beyond this value ([long]::MaxValue = 9223372036854775807).