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

Posted by da_ponc on Stack Overflow See other posts from Stack Overflow or by da_ponc
Published on 2010-04-22T01:42:16Z Indexed on 2010/04/22 1:43 UTC
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powershell

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arguments

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quotes

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variables

Hi there,

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.

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