将列表连接到路径的末尾 - Powershell

Question

我有一个路径列表和一个可执行文件列表，我需要将它们连接在一起并在每个所述路径下查找每个文件。但是，当我运行我的代码时，我看到可执行文件没有被连接，只是作为一个新行添加。此外，它仅被添加到列表中的最后一个对象。请参阅下面的代码、输出和所需的输出：

代码：

   $final_paths = @();

   $paths = @("C:\Program Files\Microsoft Office\Office15", "C:\Program Files\Microsoft 
   Office Servers\OFFICE15");

   $exes = @("MSOCF.DLL", "access.exe", "word.exe", "wordCnv.exe", "WordViewer.exe", "Excel.exe", "ExcelCnv.exe", "ExcelViewer.exe", "PowerPoint.exe", 
  "PowerPointViewer.exe", "PowerPointCnv.exe", "Publisher.exe", "Project.exe", "OneNote.exe", "InfoPath.exe Groove.exe", "FrontPage.exe", 
  "SharePointDesigner.exe", "Visio.exe", "VisioViewer.exe", "Lync.exeOutlook.exe", "WINPROJ.EXE")

  foreach ($exe in $exes)
  {
    $final_paths += $paths"$exe";
    write-output $final_paths;
  }

  foreach ($found_path in $final_paths)
  {
    $file = Get-Item -Path $found_path -ErrorAction Ignore;
    Write-Output $file

示例输出：

C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
Excel.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
ExcelCnv.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
ExcelViewer.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
PowerPoint.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
PowerPointViewer.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
PowerPointCnv.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
Publisher.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
Project.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
OneNote.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
InfoPath.exe Groove.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
FrontPage.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
SharePointDesigner.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
Visio.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
VisioViewer.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
Lync.exeOutlook.exe
C:\Program Files\Microsoft Office\Office15        
C:\Program Files\Microsoft Office Servers\OFFICE15
WINPROJ 
...
...
...
you get the idea

期望的输出：

我只想要以字符串形式返回的文件名，如下所示：

WIINPROJ

Answer 1

为此，您可以使用 System.IO 中的 Path.Combine 方法：

$finalPaths = foreach($path in $paths)
{
    foreach($exe in $exes)
    {
        Get-Item ([System.IO.Path]::Combine($path, $exe)) -EA Ignore
    }
}

$finalPaths.Name # => Should be the Names of the files found

或者，作为Lee_Dailey suggested in his comment, you can use Join-Path实现相同的结果：

$finalPaths = foreach($path in $paths)
{
    foreach($exe in $exes)
    {
        Get-Item (Join-Path $path -ChildPath $exe) -EA Ignore
    }
}

Answer 2

为什么要尝试将路径连接成全名，然后尝试使用 Get-Item 确定该文件是否存在？

使用Get-ChildItem事情会容易很多，因为

它可以处理数组 -Path 参数中的文件夹路径
您只会获得针对通过 Where-Object 子句过滤的特定文件返回的 FileInfo 对象，因此您之后无需检查它们是否存在
你得到的信息不仅仅是文件名，所以你可以稍后决定输出中你想要的是什么

$paths = 'C:\Program Files\Microsoft Office\Office15', 'C:\Program Files\Microsoft Office Servers\OFFICE15'
$exes  = 'MSOCF.DLL', 'access.exe', 'word.exe', 'wordCnv.exe', 'WordViewer.exe', 'Excel.exe', 'ExcelCnv.exe', 'ExcelViewer.exe', 'PowerPoint.exe', 
         'PowerPointViewer.exe', 'PowerPointCnv.exe', 'Publisher.exe', 'Project.exe', 'OneNote.exe', 'InfoPath.exe Groove.exe', 'FrontPage.exe', 
         'SharePointDesigner.exe', 'Visio.exe', 'VisioViewer.exe', 'Lync.exeOutlook.exe', 'WINPROJ.EXE'

$files = Get-ChildItem -Path $paths -File | Where-Object { $exes -contains $_.Name }

现在您决定要从 $files 集合

中检索什么属性

# just the Name?
$files.Name | Select-Object -Unique  # unique in case you found the same exe in both folders

# the Fullname perhaps?
$files.FullName

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