How to capture the exception raised in the scriptblock of start-job?

Question

I have the following script,

$createZip = {
    Param ([String]$source, [String]$zipfile)
    Process { 
        echo \"zip: $source`n     --> $zipfile\"
         


        

Answer 1

I was able to "rethrow" the exception in the main thread by using:

Receive-Job $job -ErrorAction Stop

I'll my use case as an example. It can easily be applied to the OP.

$code = {
    $Searcher = New-Object -ComObject Microsoft.Update.Searcher
    #Errors from Search are not terminating, but will be present in the output none the less.
    $Results = $Searcher.Search('IsInstalled=0  and IsHidden=0')
    $Results.Updates
};
$job = Start-Job -ScriptBlock $code;
$consume = Wait-Job $job -Timeout 600;

if ($job.state -eq 'Running') {
    Stop-Job $job
    throw 'Windows update searcher took more than 10 minutes. Aborting' 
};

#Captures and throws any exception in the job output
Receive-Job $job -ErrorAction Stop;
Write-Host "Finished with no errors"; #this will not print if there was an error

Works in v2.0.

Note that if the error within the job is non-terminating, the subsequent lines will continue to execute. But, this will not be obvious in the output returned from Receive-Job, as Receive-Job "terminates half way thorugh" - it throws out of it's self when the error object is encountered.

One way to avoid that is to wrap the whole block in a try {} catch{throw;}

Also, Job state will not be 'Failed' if the exception is non-terminating