powershell - 复制项是否更改PowerShell中目标参数的类型

Question

我希望这是一个愚蠢的错误，而我却忽略了一些非常简单的事情。我具有映射网络驱动器并将网络驱动器的内容复制到目标位置的功能。最后，我返回目标路径以供以后使用。但是，它似乎正在为目标路径返回不同类型的对象。以下是代码段:

  function CopyDropFolder {
 param(
    [string] $dropFolder,
    [string] $releaseName,
    [string] $mapDrive
 )

 $stageDirectory= $('c:\temp\' + $releaseName + '-' + (Get-Date -Uformat %Y%m%d-%H%M).ToString() + '\')
 [string]$destinationDirectory = $stageDirectory
 Write-Host 'Mapping Folder ' $dropFolder ' as '  $mapDrive 
 MountDropFolder -mapfolder $dropFolder -mapDrive $mapDrive

 $sourceDir = $mapDrive + ':' + '\'  
 Write-Host 'Copying from mapped drive (' $sourceDir ') to ' $stageDirectory
 Copy-Item  $sourceDir -Destination $stageDirectory -Recurse
 Write-Host $destinationDirectory
 return $destinationDirectory  
 }

我将函数调用如下:

$stageDirectory = CopyDropFolder -dropFolder $mapFolder -releaseName $releaseName -mapDrive $newDrive
Write-Host 'Staged to ' $stageDirectory 

函数(Write-Host $ destinationDirectory)的输出为:

c:\temp\mycopieddirectory-20161228-1422\

但是，从进行调用的主脚本中，输出为:

Staged to  Z c:\temp\mycopieddirectory-20161228-1422\

似乎返回的stageDirectory变量以某种方式映射到Z ::这是在函数内映射的新驱动器。

关于如何只返回函数中上面打印的路径的任何想法？

Answer 1

PowerShell具有管道的概念。调用的所有返回值的值都不会分配给变量或管道e。 G。该函数将返回Out-Null cmdlet(即使您未明确使用return关键字)。因此，您应该将函数内的输出通过管道传递给Out-Null:

 function CopyDropFolder {
 param(
 [string] $dropFolder,
 [string] $releaseName,
 [string] $mapDrive
 )

 $stageDirectory= $('c:\temp\' + $releaseName + '-' + (Get-Date -Uformat %Y%m%d-%H%M).ToString() + '\')
 [string]$destinationDirectory = $stageDirectory
 Write-Host 'Mapping Folder ' $dropFolder ' as ' $mapDrive 
 MountDropFolder -mapfolder $dropFolder -mapDrive $mapDrive | Out-Null

 $sourceDir = $mapDrive + ':' + '\' 
 Write-Host 'Copying from mapped drive (' $sourceDir ') to ' $stageDirectory
 Copy-Item $sourceDir -Destination $stageDirectory -Recurse | Out-Null
 Write-Host $destinationDirectory
 return $destinationDirectory 
 }

另外，您可以像这样重构您的方法:

function Copy-DropFolder 
 {
     [CmdletBinding()]
     param
     (
        [string] $dropFolder,
        [string] $releaseName,
        [string] $mapDrive
     )

     $stageDirectory = Join-Path 'c:\temp\' ('{0}-{1}' -f $releaseName, (Get-Date -Uformat %Y%m%d-%H%M).ToString())

     MountDropFolder -mapfolder $dropFolder -mapDrive $mapDrive | Out-Null
     Copy-Item "$($mapDrive):\"  -Destination $stageDirectory -Recurse | Out-Null

     $stageDirectory
 }

三个主要改进:

使用批准的动词(Copy-DropyFolder)

使用Join-Path cmdlet

删除了Write-Host输出(您会发现很多文章，为什么不应该使用Write-Host)。