springboot命令启动

Posted 2020-09-22 随笔

gradle打jar包命令

jar {
    doFirst {
        def jarFiles = \'\';
        configurations.compile.collect {
            jarFiles += it.name + " "
        }
        manifest {
            attributes(
                    "Manifest-Version": "1.0",
                    "Class-Path": jarFiles,
                    "Main-Class": "com.test.TestApplication")
        }
    }
}

task deploy(type: Copy) {
    doFirst {
        println "cmd /c gradle clean".execute().text
        println "cmd /c gradle build -x test".execute().text
    }

    from configurations.runtime
    into "../jar"

    from "build/libs/test-1.0-SNAPSHOT.jar"
    into "../jar"
}

发布linux目录结构

 

 1.jar存放所有jar文件

2.application.yml配置文件

server:
  address: 192.168.1.100 
  port: 10001

3.logs.out日志文件,无内容

4.web配置文件

#!/bin/bash
case $1 in 
    start)   
        java -Dfile.encoding=UTF-8 -jar jar/test-1.0-SNAPSHOT.jar --spring.config.location=application.yml > logs.out & 
        ;;
    stop)    
        ps -ef|grep jar/test-1.0-SNAPSHOT* |grep -v grep |awk \'{print $2}\'  | sed -e "s/^/kill -9 /g" | sh - 
        ;;
    restart)
        "$0" stop
        sleep 3
        "$0" start
        ;;
    status)  ps -ef|grep jar/test-1.0-SNAPSHOT*
        ;;
    *)       
        echo "Example: lpservice.sh [start|stop|restart|status]" ;;
esac

启动项目直接使用脚本启动  sh web start && tail -200f logs.out

在windows下面运行 需要加一个 .bat 文件，文件内容如下，写完后直接双击 .bat 文件就可以启动项目

chcp 65001
java -jar -Dfile.encoding=UTF-8 jar/test-1.0-SNAPSHOT.jar --spring.redis.database=10 --server.port=8088 --server.address=test.com --test.staticPath=file:../code/