Capturing standard out and error with Start-Process

Question

Is there a bug in PowerShell's Start-Process command when accessing the StandardError and StandardOutput properties?

If I run the following I get no output:

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

But if I redirect the output to a file I get the expected result:

$process = Start-Process -FilePath ping -ArgumentList localhost -NoNewWindow -PassThru -Wait -RedirectStandardOutput stdout.txt -RedirectStandardError stderr.txt

Answer 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