I have a list of directories with numbers. I have to find the highest number and and increment it by 1 and create a new directory with that increment value. I am able to sor
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
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
).
Use:
$filelist = @(11, 1, 2)
$filelist | sort @{expression={$_[0]}} |
% {$newName = [string]([int]$($_[0]) + 1)}
New-Item $newName -ItemType Directory
Once you have selected the highest value, which is "12" in my example, you can then declare it as integer and increment your value:
$FileList = "1", "2", "11"
$foldername = [int]$FileList[2] + 1
$foldername
Use:
$filelist = @("11", "1", "2")
$filelist | sort @{expression={[int]$_}} | % {$newName = [string]([int]$_ + 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.