Parse env:Path with PowerShell

The environment variable PATH (combined SYSTEM PATH and USER PATH) can be tricky to parse if you want to check for each folder it contains. This is due to the support of various formatting styles. For example, this is a valid PATH statement:

C:WINDOWS;”C:Path with semicolon; in the middle”;”E:Path with semicolon at the end;”;;C:Program Files;

Notice the differences

Folder with spaces
Folder without spaces
Folder with an ending backslash ( )
Folder without an ending backslash ( )
Folder with a semicolon (; ) in the middle
Folder with a semicolon (; ) at the end
Blank folder (;; )

A few references can be found…

But I could find no fully working method to parse this complexity, so I wrote a PowerShell Function to handle it.

Function Get-ENVPathFolders { #.Synopsis Split $env:Path into an array #.Notes # - Handle 1) folders ending in a backslash 2) double-quoted folders 3) folders with semicolons 4) folders with spaces 5) double-semicolons I.e. blanks # - Example path: 'C:WINDOWS;"C:Path with semicolon; in the middle";"E:Path with semicolon at the end;";;C:Program Files;' # - 2018/01/30 by [email protected] - Created $PathArray = @() $env:Path.ToString().TrimEnd(';') -split '(?=["])' | ForEach-Object { #remove a trailing semicolon from the path then split it into an array using a double-quote as the delimiter keeping the delimiter If ($_ -eq '";') { # throw away a blank line } ElseIf ($_.ToString().StartsWith('";')) { # if line starts with "; remove the "; and any trailing backslash $PathArray += ($_.ToString().TrimStart('";')).TrimEnd('') } ElseIf ($_.ToString().StartsWith('"')) { # if line starts with " remove the " and any trailing backslash $PathArray += ($_.ToString().TrimStart('"')).TrimEnd('') #$_ + '"' } Else { # split by semicolon and remove any trailing backslash $_.ToString().Split(';') | ForEach-Object { If ($_.Length -gt 0) { $PathArray += $_.TrimEnd('') } } } } Return $PathArray }

To prove it out:

#output the array of PATH folders Get-ENVPathFolders #create a test file in half of the folders $i=0 Get-ENVPathFolders | ForEach-Object { $i++ If ($i%2 -eq 0) { New-Item -Path "$_" -Name $myFile -ItemType File } } #output the PATH folders with the test file Get-ENVPathFolders | ForEach-Object { If (Test-Path -Path $_$myFile) { Write-Output "Found [$_$myFile]" } }

Happy coding!