通过powershell进行Robocopy输出操作

Question

我想使用-replace删除引导路径信息。

robocopy.exe C:\FOLDER\Data2 C:\FOLDER\Data2 /l /nocopy /is /e /fp /ns /nc /njh /njs  /log:c:\temp\FolderList.txt

我的输出：

C:\FOLDER\Data2\
C:\FOLDER\Data2\Documents\
        C:\FOLDER\Data2\Documents\1.txt
        C:\FOLDER\Data2\Documents\2.txt
        C:\FOLDER\Data2\Documents\3.txt
        C:\FOLDER\Data2\Documents\4.txt
        C:\FOLDER\Data2\Documents\5.txt
C:\FOLDER\Data2\Documents\TEST\
        C:\FOLDER\Data2\Documents\TEST\5.txt

我想要的输出：

    Documents\
            Documents\1.txt
            Documents\2.txt
            Documents\3.txt
            Documents\4.txt
            Documents\5.txt
    Documents\TEST\
            Documents\TEST\5.txt

Answer 1

如果您的输出是日志文件的内容：

$path = 'C:\FOLDER\Data2\'
$pattern = [regex]::Escape($path)
$newContent = @()
Get-Content -Path "c:\temp\FolderList.txt" | ForEach-Object {$newContent += $_ -replace $pattern, ''}
Set-Content -Path "c:\temp\FolderList.txt" -Value $newContent

如果您的输出被写入终端，您可以将robocopy的输出重定向到一个文件，对其进行读取，并用空字符串替换您不需要的路径的所有出现：

$path = 'C:\FOLDER\Data2\'
$tempFile = New-TemporaryFile
Start-Process robocopy.exe -ArgumentList "`"$path`" `"$path`" /l /nocopy /is /e /fp /ns /nc /njh /njs  /log:c:\temp\FolderList.txt" -Wait -RedirectStandardOutput ($tempFile.FullName)

$pattern = [regex]::Escape($path)
Get-Content -Path ($tempFile.FullName) | ForEach-Object {Write-Host "$($_ -replace $pattern, '')"}

Remove-Item -Path ($tempFile.FullName)