Capturing standard out and error with Start-Process

Solution 1:

That's how Start-Process was designed for some reason. Here's a way to get it without sending to file:

$pinfo = New-Object System.Diagnostics.ProcessStartInfo
$pinfo.FileName = "ping.exe"
$pinfo.RedirectStandardError = $true
$pinfo.RedirectStandardOutput = $true
$pinfo.UseShellExecute = $false
$pinfo.Arguments = "localhost"
$p = New-Object System.Diagnostics.Process
$p.StartInfo = $pinfo
$p.Start() | Out-Null
$p.WaitForExit()
$stdout = $p.StandardOutput.ReadToEnd()
$stderr = $p.StandardError.ReadToEnd()
Write-Host "stdout: $stdout"
Write-Host "stderr: $stderr"
Write-Host "exit code: " + $p.ExitCode

Solution 2:

In the code given in the question, I think that reading the ExitCode property of the initiation variable should work.

$process = Start-Process -FilePath ping -ArgumentList localhost -NoNewWindow -PassThru -Wait
$process.ExitCode

Note that (as in your example) you need to add the -PassThru and -Wait parameters (this caught me out for a while).