Process and WaitForExit instead of
I'm trying to run a program from PowerShell, wait for the exit, then get access to the ExitCode, but not having much luck. I don't want to use -Wait with Start-Process, as I need some processing to carry on in the background.
Here's a simplified test script:
cd "C:Windows"
# ExitCode is available when using -Wait...
Write-Host "Starting Notepad with -Wait - return code will be available"
$process = (Start-Process -FilePath "notepad.exe" -PassThru -Wait)
Write-Host "Process finished with return code: " $process.ExitCode
# ExitCode is not available when waiting separately
Write-Host "Starting Notepad without -Wait - return code will NOT be available"
$process = (Start-Process -FilePath "notepad.exe" -PassThru)
$process.WaitForExit()
Write-Host "Process exit code should be here: " $process.ExitCode
Running this script will cause notepad to be started. After this is closed manually, the exit code will be printed, and it will start again, without using -wait. No ExitCode is provided when this is quit:
Starting Notepad with -Wait - return code will be available
Process finished with return code: 0
Starting Notepad without -Wait - return code will NOT be available
Process exit code should be here:
I need to be able to perform additional processing between starting the program and waiting for it to quit, so I can't make use of -Wait. Any idea how I can do this and still have access to the .ExitCode property from this process?
Two things you could do I think...
Here's how you could do either:
$pinfo = New-Object System.Diagnostics.ProcessStartInfo
$pinfo.FileName = "notepad.exe"
$pinfo.RedirectStandardError = $true
$pinfo.RedirectStandardOutput = $true
$pinfo.UseShellExecute = $false
$pinfo.Arguments = ""
$p = New-Object System.Diagnostics.Process
$p.StartInfo = $pinfo
$p.Start() | Out-Null
#Do Other Stuff Here....
$p.WaitForExit()
$p.ExitCode
OR
Start-Job -Name DoSomething -ScriptBlock {
& ping.exe somehost
Write-Output $LASTEXITCODE
}
#Do other stuff here
Get-Job -Name DoSomething | Wait-Job | Receive-Job
There are two things to remember here. One is to add the -PassThru argument and two is to add the -Wait argument. You need to add the wait argument because of this defect http://connect.microsoft.com/PowerShell/feedback/details/520554/start-process-does-not-return-exitcode-property
-PassThru [<SwitchParameter>]
Returns a process object for each process that the cmdlet started. By d
efault, this cmdlet does not generate any output.
Once you do this a process object is passed back and you can look at the ExitCode property of that object. Here is an example:
$process = start-process ping.exe -windowstyle Hidden -ArgumentList "-n 1 -w 127.0.0.1" -PassThru -Wait
$process.ExitCode
# this will print 1
If you run it without -PassThru or -Wait, it will print out nothing.
Same answer here: https://stackoverflow.com/a/7109778/17822
While trying out the final suggestion above, I discovered an even simpler solution. All I had to do was cache the process handle. As soon as I did that, $process.ExitCode worked correctly. If I didn't cache the process handle, $process.ExitCode was null.
example:
$proc = Start-Process $msbuild -PassThru
$handle = $proc.Handle # cache proc.Handle http://stackoverflow.com/a/23797762/1479211
$proc.WaitForExit();
if ($proc.ExitCode -ne 0) {
Write-Warning "$_ exited with status code $($proc.ExitCode)"
}