Cannot bind argument to parameter 'Path' it is null and cannot find path

Question

Trying to get these two to work but keep getting errors. Basically looking to pickup all files from C:\Temp\Test with .txt extension and copy it to Server1 and Server2 D:\Temp\Test.

Doesn't work...

$servers = "Server1","Server2"
$SourcePath = (Get-ChildItem C:\Temp\Test *.txt).Name
$servers | ForEach {
Invoke-Command $servers -ScriptBlock {
$CompName = (Get-WmiObject -Class Win32_ComputerSystem).Name
$DestPath = "\\$CompName\D$\Temp\Test"
Copy-Item $SourcePath -Destination $DestPath -Recurse
}
}

Hint: check output from `$servers | ForEach { "a server = $_"; $servers }`. Moreover, if your script throws an exception, please [edit] your script to include the exception, with both the message and the stack trace. "*Doesn't work...*" enunciation does not suffice! — JosefZ, Nov 29 '16 at 21:32
Possible duplicate of [How can I pass a local variable to a script block executed on a remote machine with Invoke-Command?](http://stackoverflow.com/questions/35492437/how-can-i-pass-a-local-variable-to-a-script-block-executed-on-a-remote-machine-w) — user4003407, Nov 29 '16 at 23:15

score 2 · Answer 1 · answered Nov 29 '16 at 21:19

This is a common mistake actually. When you use Invoke-Command to invoke your scriptblock on the remote server it creates a new instance of PowerShell on that remote computer. That new instance of PowerShell has no idea what the $SourcePath variable is, since it was never set in that new instance. To work around this give your scriptblock a parameter, and then supply the value of $SourcePath when you invoke to scriptblock. It can be done like this:

$servers = "Server1","Server2"
$SourcePath = (Get-ChildItem C:\Temp\Test *.txt).Name
$servers | ForEach {
    Invoke-Command $servers -ScriptBlock {
        Param($SourcePath)
        $CompName = (Get-WmiObject -Class Win32_ComputerSystem).Name
        $DestPath = "\\$CompName\D$\Temp\Test"
        Copy-Item $SourcePath -Destination $DestPath -Recurse
    } -ArgumentList $SourcePath
}

Cannot bind argument to parameter 'Path' it is null and cannot find path

1 Answers1