startup.bat @echo off set path=X:\xxxxxxx\Java\JDK\jre\bin START "项目名" "%path%\java" -jar 要启动的jar包路径 shutdown.bat 这是根据项目名停止项目运行 wmic process where (commandline LIKE "%%项目名%%" and caption="java.exe") call ter
SpringBoot项目快速启动停止脚本 1.在jar包同级目录下,创建 app.sh #!/bin/bash appName=`ls|grep .jar$` if [ -z $appName ] then echo "Please check that this script and your jar-package is in the same directory!" exit 1 fi killForceFlag=$2 function start() { count=`ps -
1.搭建一个简单的Springboot项目,最开始将启动类main函数与controller放到一个类里,可以正常启动和访问,但是将两个分开再启动时访问就会报错:This application has no explicit mapping for /error, so you are seeing this as a fallback. 2.解决方案 在main函数启动类上加注解:@SpringBootApplication(scanBasePackages = "com") 问题