将与模式匹配的行从文件夹中的所有文本文件提取到单个输出文件 [英] Extract lines matching a pattern from all text files in a folder to a single output file

问题描述

我正在尝试提取文件夹中所有文件中以"%%"开头的每一行，然后将这些行复制到单独的文本文件中.当前在PowerShell代码中使用此代码，但未得到任何结果.

I am trying to extract each line starting with "%%" in all files in a folder and then copy those lines to a separate text file. Currently using this code in PowerShell code, but I am not getting any results.

$files = Get-ChildItem "folder" -Filter *.txt
foreach ($file in $files)
{
if ($_ -like "*%%*")
{
Set-Content "Output.txt" 
}  
}

推荐答案

我认为mklement0建议使用Select-String是可行的方法.除了他的答案，您还可以将Get-ChildItem的输出传递到Select-String中，以便整个过程成为Powershell的衬里.

I think that mklement0's suggestion to use Select-String is the way to go. Adding to his answer, you can pipe the output of Get-ChildItem into the Select-String so that the entire process becomes a Powershell one liner.

类似这样的东西:

Get-ChildItem "folder" -Filter *.txt | Select-String -Pattern '^%%' | Select -ExpandProperty line | Set-Content "Output.txt"

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