Question

我有以下代码比较两个文件夹并输出与每个文件夹不同的文件，有没有办法让它输出子文件夹中文件的完整文件/文件夹路径？

干杯

 # Create varaibles to store folder paths - pass to Strings... 
 param([string]$argFolderA,[string]$argFolderB)

 # Set variables for folder
 $FolderA = Get-ChildItem -Recurse -path $argFolderA
 $FolderB = Get-ChildItem -Recurse -path $argFolderB

 # Compare the contents and output the files different within each folder. 
 Compare-Object -ReferenceObject $FolderA -DifferenceObject $FolderB >> C:\CmpOut.txt

Answer 1

使用Comapre-Object的Passthru标志，如下所示：

Compare-Object -ReferenceObject $FolderA -DifferenceObject $FolderB -Passthru | % {$_.FullName}  >> C:\CmpOut.txt

Answer 2

试试这个：

Compare-Object -ReferenceObject $FolderA -DifferenceObject $FolderB  | 
select @{n="InputObject";e={ 
     if ($_.inputobject -is [system.io.directoryinfo]) 
     {([system.io.DirectoryInfo]$_.InputObject).fullname}
      else 
      { ([system.io.FileInfo]$_.InputObject).fullname}}},SideIndicator

Answer 3

通过ForEach {$_.InputObject.FullName}管道。

像这样：

Compare-Object -ReferenceObject $FolderA -DifferenceObject $FolderB | ForEach {$_.InputObject.FullName}

Powershell - 比较文件夹内容

3 个答案: