Question

我有两个脚本可以作为单独的脚本运行，但是当我从另一个脚本中调用一个脚本时，我收到此错误消息：

异常名称：System.Management.Automation.MethodInvocationException 例外类型：异常消息：异常调用＆＃34;打开＆＃34;用＆＃34; 3＆＃34;参数：＆＃34;无法找到路径的一部分＆＃39; D：\ Data Report \＆＃39;。

被调用的脚本是将文件从＆＃34; D：\ Data Report \移动到D：\目录中的另一个文件夹。

我在运行下面的脚本时只收到此错误。有什么想法吗？

try
{
    $folder = "D:\Data Report\" # Enter the root path you want to monitor.
    $filter = '*.*'  # You can enter a wildcard filter here.
    #$scriptPath = {D:\"file_transfer.ps1"}

    # In the following line, you can change 'IncludeSubdirectories to $true if required.
    $watcher = New-Object IO.FileSystemWatcher $folder, $filter -Property @{IncludeSubdirectories = $false;NotifyFilter = [IO.NotifyFilters]'FileName, LastWrite'}

    Register-ObjectEvent $watcher Created -SourceIdentifier FileCreated -Action {
        $name = $Event.SourceEventArgs.Name
        $changeType = $Event.SourceEventArgs.ChangeType
        $timeStamp = $Event.TimeGenerated
        Write-Host "The file '$name' was $changeType at $timeStamp" -fore green

        #Call the file_transfer.ps1 script
        #Invoke-Expression $scriptPath
        Invoke-Expression D:\file_transfer.ps1
    }

} # end of try
catch  #Catch any fatel errors and send an email with description
{
    $ErrorMessage = $_.Exception.Message
    $FailedItem = $_.Exception.ItemName
    $ErrorName = $_.Exception.GetType().FullName

    $ExceptionBody = @"
    Exception Name: $ErrorName
    Exception Type: $FailedItem
    Exception Message: $ErrorMessage
    "@
    Send-MailMessage  -SmtpServer "server.com" -From "no-replies@company.com" -To "admin@company.com" -Subject "Fatel Error with Monitor Script" -Body $ExceptionBody
} # end of catch

＃

@TheMadTechnician这是另一个独立运作的脚本。

try
{
$path="D:\Data Report\" 
$Miscdir="D:\TEST\Data\Misc"
$dataFolder = "D:\Data Report\*"
$items = Get-ChildItem -Path ($path) -ErrorAction Stop
$ErrorActionPreference= 'silentlycontinue'

# enumerate the items array
foreach ($item in $items)
{

      # if the item is NOT a directory, then process it.
     if ($item.Attributes -ne "Directory")
     {
           #Write-Host $item.Name
           $filePath = $path+$item.name


}
else  #send email and exit without running.
{
    Write-Host "$filePath is a directory"
     Exit
}

function isFileLocked([string]$LockedPath) {
$file=$filePath
$oFile = New-Object System.IO.FileInfo $LockedPath
# Make sure the path is good
if ((Test-Path -Path $LockedPath) -eq $false)
{
  echo "Bad Path"
  return $false
}
      #Try opening file
  $oStream = $oFile.Open([System.IO.FileMode]::Open, [System.IO.FileAccess]::ReadWrite, [System.IO.FileShare]::None)
  if ($oStream)
  {
    echo "Got valid stream so file must not be locked"
    $oStream.Close()
    #Pick   ID from file name
    $ ID = ($item.name.Split("_")[0])
    Write-Host $ ID

        $listcsv = import-csv d:\number.csv
        foreach($number in $listcsv){

        $UNCNumber = $number.UNCNumber
        #Write-Host $UNCNumber
        ANCNumber = $number.NoseNumber
        #Write-Host ANCNumber
        ANCNumber = $number.NoseNumber

        if ($ ID -eq $UNCNumber)

   {
        #Check/create  ID folder
    ANCNumberdir="D:\TEST\Data\ANCNumber"
    Write-Host ANCNumber
    if (!(Test-Path ANCNumberdir))
     { mkdir ANCNumberdir }

     Echo "Nose number found the csv file ANCNumber" 
      $ IDdir = ANCNumber
          Write-Host " id from csv file $ ID" }
          }

#Check/create App id folder/system
   #Pick App ID from file name
    $AppID=($item.name.Split("_")[1])
    Write-Host $AppID


        $Appiddir="ANCNumberdir\$AppID"
       if (!(Test-Path $Appiddir))
     { mkdir $Appiddir 
         Write-Host $Appiddir
        }


   Move-Item ($filePath) $Appiddir
     return $false
  } # end if ($oStream)
  else 
  {

   echo "InValid stream so file is locked"
    return $true

if (!(Test-Path $Miscdir))
          { 
      mkdir $Miscdir 
      Write-Host $Miscdir
          }             
     Move-Item $path"*" $Miscdir 
  }
}# end of function
isFileLocked($filePath)

} 


# Send an email if the script ran but a file didn't get moved out of the Data Report folder    

     If (Test-Path $dataFolder)
     {
     $DataNumbers = (get-childitem $dataFolder | ?{!($_.PSIsContainer)}).Count
      }

# check how many files are in our Misc folder and email

    $MiscNumbers = (get-childitem $miscdir | ?{!($_.PSIsContainer)}).Count
    If ($MiscNumbers -gt 0)
    {
    }

    } # end of try
    catch  #Catch any fatel errors and send an email with description
    {
    $ErrorMessage = $_.Exception.Message
    $FailedItem = $_.Exception.ItemName
    $ErrorName = $_.Exception.GetType().FullName 

    $ExceptionBody = @"
    Exception Name: $ErrorName
    Exception Type: $FailedItem 
    Exception Message: $ErrorMessage 
"@
     } # end of catch

    finally
    {

    }

Answer 1

我即将发表评论，但我认为这样做会耗费太长时间，难以阅读。正如理查德所说，似乎您的问题很可能来自使用Invoke-Expression，而您只想运行其他脚本。有两种基本的方法可以做到这一点（以及更复杂的方法），所以我会快速为你运行它们。

使用呼叫操作员&运行脚本。它会这样做：

& D:\file_transfer.ps1

就这么简单。这将运行另一个脚本，它将自动运行第一个脚本。

第二种方法是点源它，这与使用调用操作符非常相似，只是它在同一个会话中运行脚本，因此其他脚本可以访问变量以及您已经声明的内容。从你的脚本看起来不需要，但它通常不会伤害我的经验。你这样做的方式是：

. D:\file_transfer.ps1

希望在不使用Invoke-Expression cmdlet的情况下清除脚本运行。

奖励信息：在回复某人的评论时，在他们的名字前放置一个@，它会向他们显示他们有消息。

Answer 2

我不知道为什么会这样做但是如果第一个脚本运行并且它在$ path中找不到文件那么它会抛出“Exception calling”Open“with”3“argument（s）：“无法找到路径的一部分'D：\ Data Report \'。”错误。所以我不知道为什么，但是当第一个脚本调用第二个脚本时，它认为它正在运行并且文件不存在。所以我输了一个： if（Test-Path -Path“$ path *”）{ 工作代码 } 否则
{ 出口 }

调用另一个脚本时出现Powershell错误

2 个答案: