从路径 PowerShell 中剥离文件

Question

I have a PowerShell Script that finds all log4j jar files on a system. I want to separate the values out of the found paths based on just path , filename , and version .

For example, if I have a returned path of C:\Users\Administrator\AppData\Roaming\.minecraft\libraries\org\apache\logging\log4j\log4j-core\2.14.1\log4j-core-2.14.1.jar , how can I get:

Just the path - C:\Users\Administrator\AppData\Roaming\.minecraft\libraries\org\apache\logging\log4j\log4j-core\2.14.1\

Just the file - log4j-core-2.14.1.jar

Just the version - 2.14.1

Answer 1

I would simply use Get-ChildItem for that.

$Item = Get-ChildItem -Path 'C:\Users\Administrator\AppData\Roaming\.minecraft\libraries\org\apache\logging\log4j\log4j-core\2.14.1\log4j-core-2.14.1.jar'

# Folder containing the file
$Item.PSParentPath

# Filename
$Item.Name

# Version extracted from the Parent path
# As string
($ITem.PSParentPath.Split('\'))[-1]
# As System.Version object
[Version]($ITem.PSParentPath.Split('\'))[-1]

Or, using Split-Path

$FullPath = 'C:\Users\Administrator\AppData\Roaming\.minecraft\libraries\org\apache\logging\log4j\log4j-core\2.14.1\log4j-core-2.14.1.jar'
$Filename = Split-Path -Path $FullPath -Leaf
$ParentPath = Split-Path -Path $FullPath -Parent
# Version from Parent path
$Version = $ParentPath.Split('\')[-1]