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How to pass a variable as an argument to a quoted command in powershell

My powershell script takes the following parameter:

Pairs ($ BackedUpFilePath)

The value that is passed to my script is:

"\ 123.123.123.123 \ Backups \ Website.7z"

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

$ WebsiteDeploymentFolder = "C: \ example"

I am trying to extract an archive using 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 the $ BackedUpFilePath to be dynamic:

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

I think I need to pass the $ BackedUpFilePath in 7z with quotes, but they seem to be removed no matter what I try. I'll quote hell.

Thanks.

EDIT: Turns out the problem was that I was navigating to "\ 123.123.123.123 \ Backups \ Website.7z". (extra single quotes)

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1 answer


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 parameters correctly.

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