Powershell循环，用于将多个txt文件拆分为单独的文件

时间：2016-08-02 18:50:27

标签： powershell powershell-v2.0 powershell-v3.0

我遇到问题，循环遍历目录中的所有文件并根据字段“[IMPORT]”将其拆分

目录：\ C \ users $ \ Pepe \ Desktop \ TestInfoFolder

目录中有以下两个文件： TestingInfo1.txt TestingInfo2.txt

他们看起来像这样：
TestingInfo1.txt 

[IMPORT]
1
2
3
4
5
6
7
8
9
10
[IMPORT]
11
12
13
14
15
16
17
18
19
20

TestingInfo2.txt 

[IMPORT]
21
22
23
24
25

所需的结果将是3个单独的文件拆分字符串“[IMPORT]”，这包含在文件中。

结果1：

[IMPORT]
1
2
3
4
5
6
7
8
9
10

结果2：

[IMPORT]
11
12
13
14
15
16
17
18
19
20

结果3：

[IMPORT]
21
22
23
24
25

当前代码：

$Path = "\\C\users$\Pepe\Desktop\TestInfoFolder"
Get-ChildItem $Path | foreach-object {
   $InputFile = $_.FullName
   $Reader = New-Object System.IO.StreamReader($InputFile)
   $N = 1

While (($Line = $Reader.ReadLine()) -ne $null) {
   if ($Line -match "[IMPORT]") {
       $OutputFile = $matches[0] + $N + ".txt"
       $N++
   }

   Add-Content (Join-Path $Path $OutputFile) $Line
   }
}

当前查询的问题是它将ResultsInfo2.txt的结果附加到结果1的结果中。如果您需要进一步说明，请与我们联系。

3 个答案:

答案 0 :(得分：1)

正则表达式是你的朋友。这应该为$ Path中的所有文件提供IMPORT序列的所有实例：

ls $Path | cat | select-string -AllMatches "(\[IMPORT\][\s*\d+]+)" | % {$_.Matches.Value}

答案 1 :(得分：1)

我会做类似于Jon Dechiro的事情，但我会将所有文本作为多行字符串收集到一个变量中，然后将其拆分为包含从[IMPORT]到下一个FOR的所有内容的块（或者文件的结尾），然后输出$Path = "C:\Users\Pepe\Desktop\TestInfoFolder" $InputText = (Get-Content "$Path\*.txt" -Raw) -join "`r`n" $SplitText = $InputText -split "(?s)(\[IMPORT].*?)(?=\[IMPORT]|$)"|?{$_} New-Item -Path $Path\Output -ItemType Directory -Force|Out-Null For($i=0;$i -le $SplitText.Count;$i++){ $SplitText[$i] | Set-Content "$Path\Output\Results$($i+1).txt" -Force } 循环中的每个项目：

<div class="ui-grid-viewport ng-isolate-scope" ng-style="colContainer.getViewportStyle()" role="rowgroup" ui-grid-viewport="" style="overflow-x: hidden; overflow-y: scroll;">
<div class="ui-grid-canvas">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index" style="margin-top: 630px;">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index" style="">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   <div class="ui-grid-row ng-scope" ng-style="Viewport.rowStyle(rowRenderIndex)" ng-repeat="(rowRenderIndex, row) in rowContainer.renderedRows track by $index">
   </div>
</div>

这将创建一个文件夹＆＃34;输出＆＃34;在路径中，并将结果保存在那里。

答案 2 :(得分：0)

我会这样做：

$SourcePath = "C:\Users\jon_dechiro\Desktop\Test"
$OutputPath = "C:\Temp"
$Count = 1
foreach ($File in (Get-ChildItem -Path $SourcePath -File))
{
    $FileRawContents = Get-Content $File.FullName -Raw
    foreach ($Import in ($FileRawContents -split '\[IMPORT\]'))
    {
        if ($Import -match '\d')
        {
            $ImportEdited = $Import -replace '\n', "`r`n"
            "[IMPORT]$ImportEdited" | Out-File "$OutputPath\Results$Count.txt"
            $Count++
        }
    }
}

打开每个文件并按[IMPORT]拆分，然后按照指示将每个结果发送到输出文件。当然，将$ SourcePath和$ OutputPath变量更改为您需要的。

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