At a loss with Get-ChildItem behavior

This topic contains 2 replies, has 2 voices, and was last updated by  Bard 4 years, 5 months ago.

  • Author
  • #12801



    I've been at this for a really long time and I'm at my wits end. My script is below. I'm trying to figure out why I'm not getting any objects returned from line 66 when I specify multiple filters in the -include parameter. The script works as expected if I only provide one item in the include filter. If I run the get-childitem cmdlet via my output from line 64 I get the expected number of objects back. Can someone please let me know why the behavior differs when I use the cmdlet in my script versus run get-childitem cmdlet outside my script?

    Function Copy-WebLogs
    param (

    $UNIT_TEST = @{servers="";
    switch ($servertype)
    'wsvcctt' {
    $hash = $WEB_SERVICE_CTT

    'unittest' {
    $hash = $UNIT_TEST


    Process {
    $webs= $hash.servers

    switch ($logType) {
    'Tomcat' {
    $logDirectory = $hash.Tomcat

    'IIS' {
    $logDirectory = $hash.IIS

    'AppLogs' {
    $logDirectory = $hash.AppLogs

    < #Check to see if the $target path exists, if not create it. #>
    If (!(test-path -Path $target)) {
    New-Item -Path $target -ItemType Directory

    foreach($web in $webs)

    If (!(test-path -Path $target\$web))
    New-Item -Path $target\$web -ItemType Directory -ErrorAction Stop

    #below is line 68
    "Get-ChildItem -path \\$web\$logDirectory\* -Include $logName"
    #Below is line 70
    Get-ChildItem -path \\$web\$logDirectory\* -Include $logName | ForEach-Object {"Found stuff"}

  • #12805

    Dave Wyatt

    What's in your $logName variable when it works, an when it does not?

    I suspect that the problem is that you've declared the $logName parameter as a [string], not [string[]]. When you want to pass multiple multiple values to the -Include parameter of Get-ChildItem, you do so by passing it an array of strings. Even if you're passing an array to your Copy-WebLogs function, PowerShell is converting that into a single String object because of how you declared $logName in your param block (which typically means you'll have a space-separate list, as seen here:)

    $variable = [string]('One','Two','Three','Four','Five')
    < #
    One Two Three Four Five
  • #12846


    Thanks Dave I really appreciate it. The problem was with my declaration of $logName – as you suspected.

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