Question

感谢您关注我的问题。

我在给定目录中有五个配置文件。在我的批处理脚本中，我想读取这些文件名并将其提示给用户。用户选择配置文件后，从中读取变量。

任何人都可以在这里帮助我一些逻辑。所以，我可以把它推进去。

谢谢。

Answer 1

这样的批处理或.cmd文件演示了菜单技术（没什么特别的，用户必须准确输入文件名）。关键项目：

FOR
SET /P
IF EXIST

祝你好运！

@echo off

REM Show the user the list and ask them which one to use
echo.
echo Please select one of:
echo.
for %%F in ("D:\A Given Directory\*.config") do echo     %%~nxF
echo.
set SEL_CFGFNM=
set /P SEL_CFGFNM=Which configuration file: 

REM Make sure they answered, and that the file exists
if "%SEL_CFGFNM%" == "" goto ENDIT
if NOT EXIST "D:\A Given Directory\%SEL_CFGFNM%" goto NOCFG

REM User has selected file "D:\A Given Directory\%SEL_CFGFNM%" and it exists
REM Do whatever you want to do with that file now

REM Don't fall through the exit messages
goto ENDIT

REM Exit Messages
:NOCFG
echo.
echo ERROR:  Configuration file "%SEL_CFGFNM%" is not on the list
echo.
goto ENDIT

REM Cleanup
:ENDIT
set SEL_CFGFNM=

批处理脚本读取目录中的文件名并要求用户选择一个

1 个答案: