将结果导出到 CSV 时出现系统对象错误

System Object Error when exporting results to CSV

我正在尝试将我的结果从比较对象导出到 csv,但在导出时出现错误。当我在 excel 中调用它时看起来还不错。我的猜测是,只要有多个值的输出,就会放置错误而不是值。

这是我的 csvs past.csv

VKEY
V-12345
V-23456
V-1111

current.csv

VKEY
V-12345
V-6789
V-23456
V-256

我的新 csv 应该是

Past, Current
V-6789,V-1111
V-256

我现在得到的是

Past, Current
System.Object[],@{vkey=V-1111}

.

$Past = Import-CSV "past.csv"
$Current = Import-CSV "Current.csv"

$pastchange = Compare-Object $Past $Current -Property vkey | Where-Object {$_.SideIndicator -eq '=>'} | Select-Object VKEY
$currentchange = Compare-Object $Past $Current -Property vkey | Where-Object {$_.SideIndicator -eq '<='} | Select-Object VKEY

$obj = New-Object PSObject
$obj | Add-Member NoteProperty Past $pastchange
$obj | Add-Member NoteProperty Current $currentchange
$obj | Export-Csv "ChangeResults.csv" -NoTypeInformation

$obj.Past 列中显示的 System.Object[] 只是一组自定义对象,类似于 $obj.Past 列中的 @{vkey=V-1111}证明:

PS D:\PShell> $obj
$obj.Past.Gettype() | Format-Table
$obj.Current.Gettype()
"---"
$obj.Past | ForEach-Object { $_.Gettype() }

Past                                       Current                                  
----                                       -------                                  
{@{vkey=V-6789}, @{vkey=V-256}}            @{vkey=V-1111}                           

IsPublic IsSerial Name                                     BaseType                 
-------- -------- ----                                     --------                 
True     True     Object[]                                 System.Array             

IsPublic IsSerial Name                                     BaseType                 
-------- -------- ----                                     --------                 
True     False    PSCustomObject                           System.Object            
---
True     False    PSCustomObject                           System.Object            
True     False    PSCustomObject                           System.Object            

我的解决方案使用 ArrayList Class (.NET Framework):

$csvOutFile = "d:\test\ChangeResults.csv"    # change to fit your circumstances
$PastInFile = "d:\test\past.csv"
$CurrInFile = "d:\test\curr.csv"

$Past = Import-CSV $PastInFile
$Curr = Import-CSV $CurrInFile

# compare CSV files and convert results to arrays
$PastCh=@(,                                     <# always return an array           #>
            $( Compare-Object $Past $Curr -Property vkey | 
                 Where-Object { $_.SideIndicator -eq '=>' } ) |
             ForEach-Object { $_ | Select-Object -ExpandProperty vkey }
          )
$CurrCh=@(,                                     <# even if no SideIndicator matches #>
            $( Compare-Object $Past $Curr -Property vkey | 
                 Where-Object { $_.SideIndicator -eq '<=' } ) |
             ForEach-Object { $_ | Select-Object -ExpandProperty vkey }
          )

[System.Collections.ArrayList]$csvout = New-Object System.Collections.ArrayList($null)
$auxHash = @{}                                       # an auxiliary hash object

$max = ($CurrCh.Count, $PastCh.Count | Measure-Object -Maximum).Maximum
for ($i=0; $i -lt $max; $i++) {
    Try { $auxHash.Past = $PastCh.GetValue($i) } Catch { $auxHash.Past = '' }
    Try { $auxHash.Curr = $CurrCh.GetValue($i) } Catch { $auxHash.Curr = '' }
    $csvout.Add((New-Object PSObject -Property $auxHash)) > $null
}
$csvout | Format-Table -AutoSize      #  show content: 'variable $csvout'

$csvout | Export-Csv $csvOutFile -NoTypeInformation 
Get-Content $csvOutFile               #  show content: "output file $csvOutFile"

输出:

PS D:\PShell> D:\PShell\SO753277.ps1

Past   Curr  
----   ----  
V-6789 V-1111
V-256        


"Past","Curr"
"V-6789","V-1111"
"V-256",""

PS D:\PShell> 

这是 TryCatch 块的替代方案:

    <# another approach instead of `Try..Catch`:
    if ($i -lt $PastCh.Count) { $auxHash.Past = $PastCh.GetValue($i)
                       } else { $auxHash.Past = '' }
    if ($i -lt $CurrCh.Count) { $auxHash.Curr = $CurrCh.GetValue($i)
                       } else { $auxHash.Curr = '' }
    #>