PowerShell：将每行 CSV 导出到单个文件

Question

我有一个制表符分隔的 CSV，需要将其拆分为每一行的单独 CSV，文件名为 [FirstName] [LastName].csv。

此外，我需要处理一些数据，特别是从 SSN 中删除破折号。

示例内容

FirstName  LastName   SSN
=========  =========  ======
Albert     Abernathy  111-11-1111
Billy      Barty      222-22-2222
Chris      Canton     333-33-3333

所需的输出（文件名 | CSV 内容）

Albert Abernathy.csv | Albert     Abernathy  111111111
Billy Barty.csv      | Billy      Barty      222222222
Chris Canton.csv     | Chris      Canton     333333333

这是我目前所拥有的。

$csvFile = "C:\Test\List.csv"
$folderPath = Split-Path -Parent $csvFile
$csv = Import-Csv -Path $csvFile -Delimiter "`t" 
| Where-Object {$_.LastName -ne ""} ##Prevent importing blank lines
| Foreach-Object {$_.SSN -Replace "-","" } ##Remove dashes from SSN
foreach ($line in $csv)
{   $saveName = $line.FirstName + " " + $line.LastName + ".csv"
    Export-Csv -Delimiter "`t" -Path $folderPath\$saveName
}

到目前为止，所有这些都运行了，但它只是在最后一个子句之后暂停，没有任何反应。当我从谷歌搜索中拼凑出来时，我确定我的语法需要改进，所以如果有人能指出我正确的方向，我将不胜感激

--更新--

感谢 alroc 为我指明了正确的方向，感谢 mjolinor 提供了我在其他任何地方都找不到的宝贵的 REPLACE 语法。这是我的最终脚本。

$csvFile = "C:\Test\List.csv"
$folderPath = Split-Path -Parent $csvFile

### filter out blank records and delete dashes from SSN ###
(Get-Content $csvFile | 
    Select-Object |
    Where-Object {$_.LastName -ne ""}) | #Rows with non-blank last name
    Foreach-Object {$_ -Replace '(\d{3})-(\d{2})-(\d{4})','$1$2$3'} |
    Set-Content $csvFile

### import csv data and export file for each row ###
$csv = Import-Csv -Path $csvFile -Delimiter "`t"
foreach ($row in $csv) {
    $outPath = Join-Path -Path $folderPath -ChildPath $($row.FirstName + " " + $row.LastName + ".csv");
    $row | Select-Object | 
        ConvertTo-Csv -NoTypeInformation -Delimiter "`t" | 
        Foreach-Object {$_.Replace('"','')} | #Delete quotes around each "cell"
        Out-File $outPath
    (Get-Content $outPath | 
        Select-Object -Skip 1) | #Delete header
        Set-Content $outPath
}

Answer 1

我不清楚您所说的“它只是在最后一个子句之后暂停并且什么都没有发生”是什么意思，但这可能与您将 import-csv 传递给您的其余流程并尝试分配结果有关到$csv。这对我有用：

$csvFile = "C:\Test\List.csv"
$folderPath = Split-Path -Parent $csvFile
$csv = Import-Csv -Path $csvFile -Delimiter "`t" 
foreach ($record in $csv) {
    $OutFileName = join-path -Path $folderPath -ChildPath $($record.FirstName + " " + $record.LastName + ".csv");
    $record|select-object FirstName,LastName,@{name="SSN";expression={$_.SSN -replace "-",""}}|export-csv -NoTypeInformation -Path $OutFileName -Delimiter "`t";
}

或者，如果您想在没有中间变量的情况下执行此操作，请保持更多流水线：

$csvFile = "C:\Test\List.csv"
$folderPath = Split-Path -Parent $csvFile
Import-Csv -Path $csvFile -Delimiter "`t" | foreach {
    $OutFileName = join-path -Path $folderPath -ChildPath $($_.FirstName + " " + $_.LastName + "2.csv");
    $_|select-object FirstName,LastName,@{name="SSN";expression={$_.SSN -replace "-",""}}|export-csv -NoTypeInformation -Path $OutFileName -Delimiter "`t";
}

Answer 2

FWIW - 未经测试。尽可能多地保留原始脚本。

$csvFile = "C:\Test\List.csv"
$folderPath = Split-Path -Parent $csvFile.fullname

Import-Csv -Path $csvFile -Delimiter "`t" |
 Where-Object {$_.LastName -ne ""} |  ##Prevent importing blank lines
 Foreach-Object {
  $_.SSN = $_.SSN -Replace "-",""  ##Remove dashes from SSN
  $saveName = $_.FirstName + " " + $_.LastName + ".csv" 
  Export-Csv $_ -Delimiter "`t" -Path $folderPath\$saveName
  }