批处理-将以模式开头的文件移动到某个文件夹 [英] Batch - Move file that start with pattern to a certain folder

问题描述

我有一个这样组织的文件列表:test％MM％YYYY％DD.txt，例如:

I have a list of files organized like this: test%MM%YYYY%DD.txt, for example:

test01201401.txt
test01201402.txt
test01201403.txt
...
test02201401.txt
test02201402.txt
...

我想创建像\test%MM%YYYY这样的基于月度的文件夹(例如\test012014或\test022014)，然后将所有基于每日的.txt文件移动到相应的文件夹中，例如，所有test012014*文件都被移动了到\test012014文件夹，所有test022014*文件都移到\test022014文件夹，依此类推. 谢谢！

I would like to create monthly-based folders like \test%MM%YYYY (e.g. \test012014 or \test022014) and then move all the daily based .txt files into the respective folder, for example all test012014* files are moved to the \test012014 folder and all test022014* files are moved to the \test022014 folder and so on. Thanks!

推荐答案

@echo off
setlocal enabledelayedexpansion
for /f %%f in ('dir /b ^| findstr /r "^test[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]\.txt$"') do (
    set "filename=%%~nf"
    if not exist "!filename:~0,10!" md "!filename:~0,10!"
    move "%%~f" "!filename:~0,10!"
)

对于与该正则表达式^test[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]\.txt$匹配的每个文件名(以test开头，后跟8位数字并以.txt结尾的文件名)，它将检查其名称是否与文件名的前10个字符匹配的文件夹(t e s t M M Y Y Y Y)已经存在，如果没有创建它，那么它将文件移到那里.

For every filename that matches this regex ^test[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9]\.txt$ (a filename starting with test, followed by 8 digits, and ending with .txt), it checks if a folder whose name matches the first 10 characters of the filename (t e s t M M Y Y Y Y) already exists, if not it creates it, then it moves the file there.

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