批量删除部分文件名 [英] Removing part of filename with batch

问题描述

我试图调整提供给类似问题的许多代码，但是我不认为解决方案已经发布.我的问题是，我要从中删除其余部分的部分在上一个部分之前存在2次！

I've tried to tweak a lot of the code provided to similar questions, but I don't think the solution is posted. My problem is that the part from where I want to remove the rest exists 2 times before the last!

我拥有的是一个文件夹:

What I have is a folder with:

number1-number2-number3 - some random text about the file.filetype

number1 的范围是01到99
number2 的范围为1-99999
number3 的范围为1-999，可能有2个小数，由整数与整数分隔.

number1 will range from 01 to 99
number2 will range from 1-99999
number3 will range from 1-999 with the possibility of 2 decimals, separated from whole number by .

示例文件夹c:\temp\:

15-1592-1 - file 1.doc
15-1592-2 - this is file2.pdf
15-1592-3 - this cointains subfiles.html
15-1592-3.1 - sub1.jpg
15-1592-3.2 - sub2.pdf

我需要一个文件夹，其中 number3 之后的所有内容都将从文件名中删除，而且文件类型也保持不变.

What I need is a folder where everything after the end of number3 is removed from the filename, but also the file type unaltered.

示例:

15-1592-1.doc
15-1592-2.pdf
15-1592-3.html
15-1592-3.1.jpg

我了解通过阅读所有答案，这是安静的.

I understand this is quiet possible from reading all the answers combined.

我所缺少的是编译所有知识！

What I lack is the knowledge to compile it all!

推荐答案

您要删除第一个空格之后的所有内容(不带扩展名)
如果使用修饰符(请参见for /?)，这很容易:

You want to delete everything after the first space (without the extension)
This is quite easy, if you use modifiers (see for /?):

@echo off
setlocal enabledelayedexpansion
for %%a in (??-*-*) do (
  for /f "tokens=1 delims= " %%b in ("%%~na") do (
     ECHO ren "%%a" "%%b%%~xa"
  )
)

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