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

This should be a comment really, but I don't have the reputation to leave comments.

My answer is that you should use Andy Arismendi's answer, but also output $job.ChildJobs[0].Error

As $job.ChildJobs[0].JobStateInfo.Reason.Message isn't always useful.