我有下一个批处理文件,但它按文件名匹配。这次我的列表很大,文件名没有扩展名,但是我可以忽略扩展并复制例如:filename。*到目标文件夹吗?
这是当前的脚本:
title Deploying Edithor
set src_folder=S:\ApliTelinver\Compilacion\Edithor 10.5\Pipe
set dst_folder=J:\alazarev\Objetos-Migracion-Pipe
set filelist=filelist-pipe.txt
echo Origen: %src_folder% >> "pipemigracion-!datetimef!.log"
echo Destino: %dst_folder% >> "pipemigracion-!datetimef!.log"
echo.
REM for /f %%i in (%filelist%) DO xcopy /S/E/U/Y "%src_folder%\%%i" "%dst_folder%" > "%dd%.log"
for /f "delims=" %%i in (%filelist%) do (
xcopy /S/E/U/Y "%src_folder%\%%i" "%dst_folder%" >> "pipemigracion-!datetimef!.log"
)
echo Success. >> "pipemigracion-!datetimef!.log"
echo.
echo Done - Check log pipemigracion-!datetimef!.log
echo.
pause
goto start
答案 0 :(得分:3)
以下是我对这个问题的理解。文件列表filelist-pipe.txt
包含文件名,所有名称都没有扩展名。
如果是这样,您只需要将.*
附加到XCOPY
命令中的源文件路径:
title Deploying Edithor
set src_folder=S:\ApliTelinver\Compilacion\Edithor 10.5\Pipe
set dst_folder=J:\alazarev\Objetos-Migracion-Pipe
set filelist=filelist-pipe.txt
echo Origen: %src_folder% >> "pipemigracion-!datetimef!.log"
echo Destino: %dst_folder% >> "pipemigracion-!datetimef!.log"
echo.
REM for /f %%i in (%filelist%) DO xcopy /S/E/U/Y "%src_folder%\%%i" "%dst_folder%" > "%dd%.log"
for /f "delims=" %%i in (%filelist%) do (
xcopy /S/E/U/Y "%src_folder%\%%i.*" "%dst_folder%" >> "pipemigracion-!datetimef!.log"
)
echo Success. >> "pipemigracion-!datetimef!.log"
echo.
echo Done - Check log pipemigracion-!datetimef!.log
echo.
pause
goto start
如果我还在遗漏某些东西,请告诉我。