Redirection of standard and error output appending to the same log file

Question

I need to collect the standard output and error log from several processes into one single log file.

So every output must append to this log file.

Answer 1

In order to append to a file you'll need to use a slightly different approach. You can still redirect an individual process' standard error and standard output to a file, but in order to append it to a file you'll need to do one of these things:

1. Read the stdout/stderr file contents created by Start-Process
2. Not use Start-Process and use the call operator, &
3. Not use Start-Process and start the process with .NET objects

The first way would look like this:

$myLog = "C:\File.log"
$stdErrLog = "C:\stderr.log"
$stdOutLog = "C:\stdout.log"
Start-Process -File myjob.bat -RedirectStandardOutput $stdOutLog -RedirectStandardError $stdErrLog -wait
Get-Content $stdErrLog, $stdOutLog | Out-File $myLog -Append

The second way would look like this:

& myjob.bat 2>&1 >> C:\MyLog.txt

Or this:

& myjob.bat 2>&1 | Out-File C:\MyLog.txt -Append

The third way:

$pinfo = New-Object System.Diagnostics.ProcessStartInfo
$pinfo.FileName = "myjob.bat"
$pinfo.RedirectStandardError = $true
$pinfo.RedirectStandardOutput = $true
$pinfo.UseShellExecute = $false
$pinfo.Arguments = ""
$p = New-Object System.Diagnostics.Process
$p.StartInfo = $pinfo
$p.Start() | Out-Null
$p.WaitForExit()
$output = $p.StandardOutput.ReadToEnd()
$output += $p.StandardError.ReadToEnd()
$output | Out-File $myLog -Append