服务器启动后如何使用批处理文件启动应用程序? [英] How to launch application after server startup using batch file?

问题描述

@echo off
set START_SERVER="C:\Program Files\Apache\Tomcat_6\bin\startup.bat"
call %START_SERVER%
start http://localhost:8080/MyProject/

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    我上面的脚本正在启动服务器并启动应用程序..，但两者同时工作.

    My above script is starting server and launching app.. but both working simultaneously.

    推荐答案

    您的代码正确. startup.bat完成后，start将运行. 问题可能在startup.bat中，可能是在完全启动该exe文件之前启动了一个exe文件，然后退出了. 您可以编辑startup.bat或在call行和start行之间设置timeout.

    your code is right. The startwill run, when startup.bat is finished. The problem propably is in startup.bat Propably it starts an exe-file, and then exits, before that exefile is fully started. You can edit startup.bat or set an timeoutbetween the callline and the startline.

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