且构网

分享程序员开发的那些事...
且构网 - 分享程序员编程开发的那些事

如何替换目录中每个文件的内容?

更新时间:2022-11-02 23:01:35

Get-Content "C:\Dir\*" 会给你 C:\Dir 中所有内容的内容 一次性完成,因此您将无法单独修改每个文件.C:\Dir 中的任何目录也会出错.

Get-Content "C:\Dir\*" will give you the content of everything in C:\Dir in one go, so you won't be able to modify each file individually. You'll also get errors for any directory in C:\Dir.

您需要遍历目录中的每个文件并单独处理它们:

You need to iterate over each file in the directory and process them individually:

Get-ChildItem 'C:\Dir' -File | ForEach-Object {
  $file = $_.FullName
  (Get-Content $file) -replace '1','2' | Set-Content $file
}

Get-Content 周围的括号确保在进一步处理之前读取并再次关闭文件,否则写入(仍然打开)文件将失败.

The parentheses around Get-Content ensure that the file is read and closed again before further processing, otherwise writing to the (still open) file would fail.

请注意,参数 -File 仅在 PowerShell v3 或更高版本中受支持.在旧版本中,您需要将 Get-ChildItem -File 替换为以下内容:

Note that the parameter -File is only supported in PowerShell v3 or newer. On older versions you need to replace Get-ChildItem -File with something like this:

Get-ChildItem 'C:\Dir' | Where-Object { -not $_.PSIsContainer } | ...