aloha20200628

回复 1# cckkaa888

给一个直接用递归方法 ‘截取文件名第一个多字母字段’ 的版本
以下代码存为 test.bat 置于目标文件（*.txt）所在目录下运行...

@echo off
for /f "delims=" %%a in ('dir /b/a-d *.txt') do (
	set "s=%%a"&set "$a=%%a" &setlocal enabledelayedexpansion
	set "_s="&set "_k=" & (call :getstr)
	md "!_s!" 2>nul & move /y "!$a!" "!_s!\" & endlocal
)
pause&exit/b
:getstr
	set "c=!s:~,1!" &if "!c!" geq "a" if "!c!" leq "Z" (
		set "s=!s:~1!"&set "_s=!_s!!c!"&set/a "_k+=1" & goto :getstr)
	if !_k! geq 2 (exit/b) else (set "s=!s:~1!"&set "_s="&set "_k=" & goto :getstr)
	exit/b
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aloha20200628

凡是提供正则匹配替换方法的脚本（如jscript/vbs/powershell/python/...）均可增援纯P代码以便其整个脚本代码轻量化...
以下代码调用jscript正则替换方法，存为test.bat，置于目标文件（*.txt）所在目录下运行...

@set @x=1 /* &@echo off
for /f "delims=" %%a in ('dir /b/a-d *.txt') do for /f "delims=" %%d in (
	' cscript /nologo /e:jscript "%~f0" "%%a" ') do (md "%%d" 2>nul & move /y "%%a" "%%d\")
pause&exit/b */
v = WSH.arguments; WSH.echo(v(0).match(/[a-z]{2,}/i)[0]); WSH.quit();
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