批处理脚本读取目录中的文件名并要求用户选择一个

Question

感谢您关注我的问题。

我在给定的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。一旦用户选择了一个配置文件，从它读取变量。

任何人都可以帮我在这里的一些逻辑。所以，我可以继续前进。

谢谢。

Answer 1

这样的批处理文件或.cmd这样的文件演示了菜单技术（没什么奇怪的，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=