如何返回第一个命令提示符并在bat文件中将其关闭
[英]How to go back to first command prompt and close it in bat file
问题描述
I have a bat file in which I call a couple of bat command sequentially: 
我有一个bat文件，在其中我依次调用几个bat命令：
start cmd /c a.bat 启动cmd / c a.bat
call b.bat 致电b.bat
The 1st line will run a.bat in a new prompt window, after the 2nd line finish run, I want to go back to the window started from 1st line and close it, how can I achieve that? 第一行将在新的提示窗口中运行a.bat，第二行完成运行后，我想回到从第一行开始的窗口并关闭它，我该如何实现？ I need to start a.bat in a different window, since b.bat rely on it to run, I have tried 我需要在其他窗口中启动a.bat，因为b.bat依赖于它来运行，所以我尝试了
call a.bat 呼叫a.bat
call b.bat 致电b.bat
but it never starts b.bat since a.bat starts a server and just wait. 但是它永远不会启动b.bat，因为a.bat将启动服务器并等待。 Any advice would be appreciated. 任何意见，将不胜感激。
1 个解决方案
解决方案1
0 2014-07-31 01:02:40
Use this and inside a.bat start the server with start "" "c:\\program files\\folder\\server.exe" so it will launch and return control to the a.bat and continue. 使用它并在a.bat内部使用start "" "c:\\program files\\folder\\server.exe"启动服务器，这样它将启动并将控制权返回给a.bat并继续。
call a.bat
call b.bat
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