wins中怎么用一个bat文件启动jar包和tomcat等多个服务

假设我有两个jar包,(app1jar和app2jar),有一个tomcat,有一个nginx,

为app1.jar建一个run.bat, 内容为 java -jar C:\app1\app1.jar

为app2.jar建一个run.bat, 内容为 java -jar B:\app2\app2.jar

假设各组件的具体路径如下：

• app1 放在 C 盘：C:\app1\app1.jar
• app2 放在 B 盘：B:\app2\app2.jar
• Tomcat 放在 E 盘：E:\tomcat\bin\startup.bat
• Nginx 放在 G 盘：G:\nginx\nginx.exe

然后建一个总的zongrun.bat文件,内容如下:

@echo off
echo ==============================================
echo 开始启动所有服务...
echo 启动时间: %date% %time%
echo ==============================================

:: 启动C盘的app1
echo 正在启动 C盘的app1...
start "App1 (C盘)" /d "C:\app1" cmd /k "run1.bat"

:: 启动B盘的app2
echo 正在启动 B盘的app2...
start "App2 (B盘)" /d "B:\app2" cmd /k "run2.bat"

:: 启动E盘的Tomcat
echo 正在启动 E盘的Tomcat...
start "Tomcat (E盘)" /d "E:\tomcat\bin" cmd /k "startup.bat"

:: 启动G盘的Nginx
echo 正在启动 G盘的Nginx...
start "Nginx (G盘)" /d "G:\nginx" cmd /k "nginx.exe"

echo ==============================================
echo 所有服务启动命令已发送
echo 各服务将在独立窗口中运行
echo ==============================================
pause

然后双击zongrun.bat文件,就一键启动两个jar包,一个tomcat,一个nginx了.