从 Start-Job 中获取 PID

Question

I am having trouble getting the PID from a Start-Process within a Start-Job in Powershell (latest version). The Start-Process cmdlet runs and returns the PID as expected when run outside of the Start-Job Scriptblock. When the statement is added to the Start-Job Scriptblock, no PID is returned. Can someone point this newbie in the right direction?

$myJob = Start-Job -Name newJob -ScriptBlock {
$procID = (Start-Process myprogram.exe -PassThru -ArgumentList "myArgs" -WorkingDirectory "myWorkingDir").Id
}

Answer 1

Perhaps you think that assigning to a variable in a script block being executed in a background job makes that variable visible to the caller , which is not the case.

Background jobs run in an entirely separate session, and in order to communicate data to the caller the success output stream must be used, whose content the caller must collect via the Receive-Job cmdlet:

$myJob = Start-Job -Name newJob -ScriptBlock {
 # Implicitly output the PID of then newly launched process.
 (Start-Process myprogram.exe -PassThru -ArgumentList "myArgs" -WorkingDirectory "myWorkingDir").Id
}

# Use Receive-Job to collect the background job's output, once available.
# NOTE: If `Start-Process` failed, this would result in an infinite loop.
#       Be sure to implement a timeout.
# You could use `$procID = Receive-Job -Wait -AutoRemoveJob` to wait synchronously,
# but that would defeat the purpose of a background job (see below).
while (-not ($procID = Receive-Job $myjob)) {
  Start-Sleep -Milliseconds 200
}

Taking a step back: Start-Process itself is asynchronous, so there's no need to use a background job : just start myprogram.exe directly in the caller's context:

# Launch myprogram.exe asynchronously.
# $proc receives a value once the program has *launched* and
# your script continues right after.
$proc = Start-Process myprogram.exe -PassThru -ArgumentList "myArgs" -WorkingDirectory "myWorkingDir"

# $proc.Id retrieves the PID (process ID)
# $proc.HasExited tells you whether the process has exited.
# $proc.WaitForExit() allows you to wait synchronously to wait for the process to exit.

However, if myprogram.exe is a console (terminal) application whose output you want to capture , do use Start-Job , but don't use Start-Process to launch myprogram.exe : invoke it directly from the background job:

$myJob = Start-Job -Name newJob -ScriptBlock {
  Set-Location "myWorkingDir"
  myprogram.exe
}

While that won't give you that process' ID, you can instead use the job that launched the process - $myJob - to track that particular process and its output.