Question

基本上，我有一个CSV文件，列出了要分析任何丢失文件的目录内容。

我的脚本执行此操作，但我尝试将丢失文件的结果导出到新的CSV文件。

当前脚本：

#get working directory
$documents = 'C:\Users\Me\Documents\ScriptTest'
#path to files located in the directory
$myFolder = Get-ChildItem 'C:\Users\Me\Documents\Test\1' -Recurse -Include *.txt | 
    Select-Object -ExpandProperty BaseName | 
        Where-Object{$_ -notmatch "\\FolderName\\"}
#get items located in CSV file list column
$myCSV = Import-CSV -Path 'C:\Users\Me\Documents\Test.csv' | 
    Select-Object -ExpandProperty 'File Path'

$CSVexport = '\Users\Me\Documents\Test.csv'

$results=@()
$desiredColumns = @{
        ID = $CSVexport | Select-Object ID 
        'File Path' = $CSVexport | Select-Object 'File Path'
        }

#split the CSV item names
$splits=@()
foreach($file in $myCSV){
  $split = $file.split("\\")[2]
  $split2 = $split.split(".")[0]
  $splits+=$split2
}

#compare the CSV items to the folder items
$compare = Compare-Object -ReferenceObject $splits -DifferenceObject $myFolder -includeequal

Write-Output "`n_____Results____`n"
#Write the results and export to a CSV file
$compare | ForEach-Object {
    if($_.SideIndicator -eq "<="){
        write-host "`n$($_.InputObject) is missing from the Test folder.`n"
        $results = New-Object PSObject -Property $desiredColumns
        $results.'File Path' = $_.InputObject
        $results | Export-Csv -Path C:\Users\Me\Documents\Results.csv –NoTypeInformation
    }
}

我得到的全部是：

它只获取一个丢失的文件而没有“ID”列。我希望能够获得比较结果并将其导出到包含第一个CSV文件中相关列的新CSV。例如：

我能否得到一些关于我可能做错的事情的帮助？

编辑（响应LotPings答案）

新守则：

$documents = 'C:\Users\Me\Documents'

$NewCsv = Import-CSV -Path 'C:\Users\Me\Documents\Test.csv' |
    Select-Object ID,'File Path' |  
      ForEach-Object {
        if (!(Test-Path (Join-Path $documents $_.'File Path'))){
          [PSCustomObject]@{
             ID = $_.ID
             Missing = $_.'File Path'
          }
       }
    }
    $NewCsv

输出：

Name                           Value                                                                           
----                           -----                                                                           
ID                             4                                                                               
Missing                        Test\1\File4.txt                                                                
ID                             6                                                                               
Missing                        Test\1\File6.txt

Answer 1

非常复杂。

如果你迭代你的csv，检查每个文件的存在并动态生成一个新的csv，它可以很简单：

## Q:\Test\2018\05\24\SO_50517360.ps1
$BasePath = "C:\Users\Me\Documents"

$NewCsv = Import-CSV -Path '.\Test.csv'|Select-Object ID,'File Path' |  
  ForEach-Object {
    if (!(Test-Path (Join-Path $BasePath $_.'File Path'))){
      [PSCustomObject]@{
         ID = $_.ID
         Missing = $_.'File Path'
      }
   }
}
$NewCsv

示例输出（可以通过
将其放入新文件中 $NewCsv|Export-Csv .\Missing.csv -NoTypeInformation

> .\SO_50517360.ps1

ID Missing
-- -------
4  Test\1\File4.txt
6  Test\1\File6.txt

将差异导出为CSV并与单独的CSV PowerShell中的列组合

编辑（响应LotPings答案）

1 个答案: