How to pass a variable as an argument to a command with quotes in powershell

My powershell script takes the following parameter:

Param($BackedUpFilePath)

The value that is getting passed into my script is:

"\123.123.123.123\Backups\Website.7z"

I have another variable which is the location I want to extract the file:

$WebsiteDeploymentFolder = "C:\example"

I am trying to extract the archive with the following command:

`7z x $BackedUpFilePath -o$WebsiteDeploymentFolder -aoa

I keep getting the following error:

Error: cannot find archive

The following works but I need $BackedUpFilePath to be dynamic:

`7z x '\123.123.123.123\Backups\Website.7z' -o$WebsiteDeploymentFolder -aoa

I think I need to pass $BackedUpFilePath to 7z with quotes but they seem to get stripped out no matter what I try. I am in quote hell.

Thanks.

EDIT: It turns o开发者_JAVA技巧ut the problem was I was passing in "'\123.123.123.123\Backups\Website.7z'". (extra single quotes)

The easiest way to work with external command line applications in PowerShell (in my opinion) is to use aliases. For example, the following works fine for me.

Set-Alias Szip C:\Utilities\7zip\7za.exe

$Archive = 'C:\Temp\New Folder\archive.7z'
$Filename = 'C:\Temp\New Folder\file.txt'

SZip a $Archive $Filename

PowerShell takes care of delimiting the parameters correctly.