将文件名从一个文件夹复制到另一个文件夹，同时保留原始扩展名 [英] Copy file names from one folder to another while keeping the original extensions

问题描述

我需要有关将文件名称(不是文件本身)从C驱动器复制到D驱动器的帮助.我能够在线找到以下powershell代码:

I need help with copying file names (not the files themselves) from C drive to D drive. I was able to find the following powershell code online:

$names = @()
$getPath = "C:\MyFiles"
$setPath = "D:\MyFiles"
Get-ChildItem $getPath |
    Foreach-object{
    $names += $_
}
$i = 0
Get-ChildItem $setPath |
Foreach-object{
    Rename-Item -Path $_.FullName -NewName $names[$i]
    $i++
}

严格按照相应的位置(枚举中的索引)，此代码成功地将所有文件名从C:\MyFiles重命名/复制到D:\MyFiles.

This code successfully renames/copies all file names from C:\MyFiles to D:\MyFiles, strictly by corresponding positions (indices in the enumeration).

但是，它也在更新扩展名，例如:

However, it is also updating the extensions, so for instance:

C:\MyFiles\myfile.txt将D:\MyFiles\thisfile.docx重命名为D:\MyFiles\myfile.txt

是否有一种方法可以编辑Powershell代码，使其仅重命名文件名的 base (例如myFile)，同时保留目标文件的扩展名(例如，.docx)?

Is there a way to edit the Powershell code to only rename the base of the filename (e.g., myFile) while keeping the target files' extensions (e.g., .docx)?

C:\MyFiles\myfile.txt使用

推荐答案

听起来您想根据目标目录中的相应文件重定位目标位置 中的文件源目录-保留目标目录文件的扩展名:

It sounds like you want to rename files in a target directory positionally, based on corresponding files in a source directory - while retaining the target directory files' extensions:

$getPath = "C:\MyFiles"
$setPath = "D:\MyFiles"
$sourceFiles = Get-ChildItem -File $getPath

$iRef = [ref] 0
Get-ChildItem -File $setPath | 
  Rename-Item -NewName { $sourceFiles[$iRef.Value++].BaseName + $_.Extension }

• 要预览生成的文件名，请将-WhatIf附加到Rename-Item调用.

  • To preview the resulting filenames, append -WhatIf to to the Rename-Item call.

    Get-ChildItem输出的[System.IO.FileInfo]对象的.BaseName属性返回不带扩展名的文件名部分.

    The .BaseName property of the [System.IO.FileInfo] objects output by Get-ChildItem returns the file-name portion without the extension.

    $_.Extension提取输入文件(即目标文件)的现有扩展名，包括前导.

    $_.Extension extracts the input file's (i.e., the target file's) existing extension, including the leading .

    请注意，传递给Rename-Item的脚本块({ ... })创建一个 child 变量作用域，因此您不能在调用方作用域中直接增加变量(它将每次使用原始值创建一个此类变量的新副本)；因此，将创建一个[ref]实例以间接 保留该数字，然后可以通过.Value属性通过该子作用域对其进行修改.

    Note that the script block ({ ... }) passed to Rename-Item creates a child variable scope, so you cannot increment a variable in the caller's scope directly (it would create a new copy of such a variable with the original value every time); therefore, a [ref] instance is created to indirectly hold the number, which the child scope can then modify via the .Value property.

    这是一个完整示例:

    ^{注意:虽然本示例使用相似的文件名和统一扩展名，但代码通常以任何名称和扩展名运行 .}

    ^{Note: While this example uses similar file names and uniform extensions, the code works generically, with any names and extensions.}

    # Determine the temporary paths.
    $getPath = Join-Path ([System.IO.Path]::GetTempPath()) ('Get' + $PID)
    $setPath = Join-Path ([System.IO.Path]::GetTempPath())  ('Set' + $PID)
    
    # Create the temp. directories.
    $null = New-Item -ItemType Directory -Force $getPath, $setPath
    
    # Fill the directories with files.
    
    # Source files: "s-file{n}.source-ext"
    "--- Source files:"
    1..3 | % { New-Item -ItemType File (Join-Path $getPath ('s-file{0}.source-ext' -f $_)) } | 
      Select -Expand Name
    
    # Target files: "t-file{n}.target-ext"
    "
    ---- Target files:"
    1..3 | % { New-Item -ItemType File (Join-Path $setPath ('t-file{0}.target-ext' -f $_)) } | 
      Select -Expand Name
    
    # Get all source names.
    $sourceFiles = Get-ChildItem -File $getPath
    
    # Perform the renaming, using the source file names, but keeping the
    # target files' extensions.
    $i = 0; $iVar = Get-Variable -Name i
    Get-ChildItem -File $setPath | 
      Rename-Item -NewName { $sourceFiles[$iVar.Value++].BaseName + $_.Extension }
    
    "
    ---- Target files AFTER RENAMING:"
    
    Get-ChildItem -Name $setPath
    
    # Clean up.
    Remove-Item -Recurse $getPath, $setPath
    

    上面的结果:

    --- Source files:
    s-file1.source-ext
    s-file2.source-ext
    s-file3.source-ext
    
    ---- Target files:
    t-file1.target-ext
    t-file2.target-ext
    t-file3.target-ext
    
    ---- Target files AFTER RENAMING:
    s-file1.target-ext
    s-file2.target-ext
    s-file3.target-ext
    

    请注意，目标文件现在如何具有源文件的基本文件名(s-file*)，但具有目标文件的原始扩展名(.target-ext).

    Note how the target files now have the source files' base file names (s-file*), but the target files' original extensions (.target-ext).

    这篇关于将文件名从一个文件夹复制到另一个文件夹，同时保留原始扩展名的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！