ScriptBlock Not Executing in PowerShell

Question

I am trying to execute an .exe file with PowerShell by doing:

$myExe = Join-Path $scriptDir "\bin\my\myProgram.exe" 
 $job = Start-Job -ScriptBlock{ 
        & $myExe }

But it doesn't execute. When in type in the file path in $myExe, the program works as expected, but when I use the above code, nothing happens. Can someone explain what I'm doing wrong; or, why I'm not getting the expected results?

To be clear, the expected result (at this point) is to get a firewall warning. Again, when I type in the path directly to the PS console, the firewall warning comes up, but using the above code, nothing seems to happen. I've also tried to check the status using Get-Process but the result is the $job variable, saying it's running.

UPDATE:

Running:

$myExe = Join-Path $scriptDir "\bin\my\myProgram.exe"
& $prndExe

Causes PowerShell to hang

UPDATE 2:

When I do the following, the program works and the process starts:

 Start-Job -ScriptBlock{ & "C:\path\Music\source\p\android\test\Automation\bin\mym\myProgram.exe"}

But using the above code doesn't. This leads me to believe I'm doing something wrong with the Start-Job cmdlet?

Have you debugged [$myExe] to make sure the full path is correct? Also, Start-Job is for background processes. Is this supposed to be a background process? — TheOptimusPrimus
– TheOptimusPrimus, Commented Jul 30, 2013 at 19:52
What is the value of $myExe when its used in Start-Job? Is it a valid path? Have you verified? — TheOptimusPrimus
– TheOptimusPrimus, Commented Jul 30, 2013 at 19:55
@TheOptimusPrimus I am writing the content of $myExe to the console, and then taking that output and putting it into the console to get it to 'work'. When I use & $myExe outside of the scriptBlock the program just hangs. — BlackHatSamurai
– BlackHatSamurai, Commented Jul 30, 2013 at 19:56
Yes, the path is valid: C:\my\Music\source\lk\android\test\Automation\bin\myfolder\myProgram.exe. I have taken that output and put it into the console, and had the expected result. Would running this as a background process make it run differently? — BlackHatSamurai
– BlackHatSamurai, Commented Jul 30, 2013 at 19:58

BlackHatSamurai · Accepted Answer · 2013-07-30 21:38:59Z

3

So the solution to this issue (one that is NOT listed in the documentation, and should be) is that if you are using variables inside ScriptBlock you have to use the ArgumentList and pass in those variables.

So to get my scriptblock working, I did:

$job = Start-Job -ScriptBlock{  
    $myExe = $args[0]

    & $myExe
} -ArgumentList @($myExe)

answered Jul 30, 2013 at 21:38

BlackHatSamurai

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3 Comments

Adi Inbar Over a year ago

I'm confused...the first line sets $myExe to a path to an exe file ($myExe = Join-Path $scriptDir "\bin\my\myProgram.exe"), and your entire script block is & $myExe...so where are the variables?

BlackHatSamurai Over a year ago

That is my point. When I set $myExe it is set outside the script block, so in order for me to use it, I have to pass it in as a parameter. Don't ask me why, but that is how it's working.

x0n Over a year ago

You have to pass it as a parameter because jobs run in a different process. The variables don't exist there.

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