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@echo off for /f "delims=" %%i in (2.txt) do set var=%%i&set str=%%i&call :lp pause>nul&start ok.txt&goto :eof :lp set a=%str:~,1% findstr "%a%" 1.txt>nul&&set flag=A set str=%str:~1% if defined str goto lp if defined flag echo %var%>>ok.txt set "flag="
@echo off&setlocal enabledelayedexpansion for /f "delims=" %%a in (2.txt) do ( set str=%%a for /f "delims=" %%i in (1.txt) do ( echo "%%i"|findstr "!str:~,1! !str:~1,1! !str:~2,1!">nul||set %%a=A ) if not defined %%a echo %%a>>ok.txt ) start ok.txt&goto :eof
Originally posted by terse at 2008-6-20 15:31: 就楼主的例子 zw19750516兄的不能删689吧 补一个 [code] @echo off&setlocal enabledelayedexpansion for /f "delims=" %%a in (2.txt) do ( set str=%%a for /f " ...
Originally posted by zw19750516 at 2008-6-20 15:58: 还是请兄弟先测试下。。。 好久不见了,兄弟一向可好?
1.txt内容: 12345 12356 01234 13456 12456 01456 2.txt内容: 123 134 156 012 013 789 689 321 ok.txt内容: 123 134 156 012 013 689 321
Originally posted by terse at 2008-6-20 16:11: zw19750516兄好 测试后是这样的 我想你在前面处理的时候,假设1.txt的第一行不包含2.txt第一行的任意一个数字 而到后面含有的时候 到最后flag还是给 ...
@echo off for /f "delims=" %%i in (2.txt) do set var=%%i&set str=%%i&call :lp pause>nul&start ok.txt&goto :eof :lp set a=%str:~,1% findstr /v "%a%" 1.txt>nul&&set /a n+=1 set str=%str:~1% if defined str goto lp if "%n%" nqu "3" echo %var%>>ok.txt set n=0