Referencing variable in scriptblock

Question

I have the following code which is parsing an XML file in PowerShell, and then iterating through the entries in the config file (which are backup jobs) and performing backups (by calling functions).

if ($xmlfile.configuration.DBAndFilesBackup.item("MSSQL").haschildnodes) {
    $xmlfile.configuration.DBAndFilesBackup.MSSQL.backup | 
        Start-RSJob -Name {$_.Name} -Throttle 2  -ScriptBlock {
            # Begin the Backup
            BeginBackup $_.Name $log

            # Backup the mssql database
            MSSQLBackup $_.Name $_.DBPath $Using:backupdir $Using:backuptempdir

            # Backup the files
            FilesBackup $_.Name $_.FolderName $_.FilesPath $Using:backupdir $Using:backuptempdir

            # End the Backup
            EndBackup $_.FolderName $_.Name $log $Using:emailTo $Using:backupdir $Using:backuptempdir
        } -FunctionsToLoad BeginBackup,MSSQLBackup,FilesBackup,EndBackup,Mailer,PrintHeader |
        Wait-RSJob |
        Receive-RSJob |
        Out-file "$ScriptDir\logs\corebackup\$ScriptName $($xmlfile.configuration.DBAndFilesBackup.MSSQL.backup.Name) $DateStamp.log"
}

Start-RSJob is a custom PowerShell module similar to the Start-Job cmdlet that handles kicking off parallel jobs.

Both RSJob and the native PowerShell Start-Job cmdlet don't seem to handle PowerShell transcription (logging). Thus I'm utilizing Write-Output, in addition to Out-File to capture the output of the jobs.

The problem I've run into is that in the Write-Output portion of the script I want to include the name of the backup in the log file name. In other words I end up with a file named "corebackup 2015-08-24.log" instead of "corebackup backupname 2015-08-24.log".

The issue is how do I pass $_.Name to Out-File. Right now a log is written, but without the job name.

Answer 1

I wasn't able to get any of the above working with throttling. In other words I could get output but then throttling broke, or i could make throttling work but not get output.

The solution I ended up using was to add a bunch of "Write-Output | Out-file $log -Append" to my script every time I want to output something.

Kinda ugly hack - but by using that inside of the scriptblock I can capture the output.

Thanks to everyone for their help and advise.