Why is my Takeown command not recognizing the path variable?

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Asked By CuriousCoder42 On

I'm having trouble with my PowerShell script. I'm using the Takeown command, but it seems like it's not picking up the variable for the file path correctly. Here's the snippet of my code:

$un = "$env:USERNAME"
$path = "C:Users%username%AppDataRoamingsomecachefolder" + $un + "controlsClientCommon.dll"
#Stop-Process -Name "SOMEPROCESS.exe", -Force
takeown /F "$path"

I was told to make sure $path is in double quotes, but that didn't help. It looks like it's searching for a literal path called $path instead of using the variable's content. Can anyone figure out how to get this working, or is this a limitation of how certain commands operate in PowerShell? By the way, when I run Write-Output $path, it shows the correct path to the file I need to access.

1 Answer

Answered By HelpfulHarry83 On

Looks like there are a couple of issues here. First off, your `$path` variable isn't being set the right way. Instead of concatenating, you can directly include the variable in the string. Also, remember that `takeown` has a character limit for paths—if it's too long, consider using `Get-Acl` and `Set-Acl` instead. Here's a revised version of your code:

```
$un = "$env:USERNAME"
$path = "C:Users$unAppDataRoamingsomecachefolder$uncontrolsClientCommon.dll"
#Stop-Process -Name "SOMEPROCESS.exe", -Force
takeown /F "$path"
```

StringSimplicity99 -

Oh, so you can actually put a variable in an existing string without needing to concatenate? That's a useful tip! Also, using Set-Acl sounds like a solid backup plan. I noticed it doesn't work well with the `%username%` syntax in paths either. Strange!

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