How to pass a variable as an argument to a command with quotes in powershell
- by da_ponc
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.
