这可以在Ubuntu中使用Systemd服务完成

[Unit]
Description=A Spring Boot application
After=syslog.target

[Service]
User=baeldung
ExecStart=/path/to/your-app.jar SuccessExitStatus=143

[Install] 
WantedBy=multi-user.target

你可以点击这个链接获得更详细的描述和不同的方法。 http://www.baeldung.com/spring-boot-app-as-a-service

Following up on Chad's excellent answer, if you get an error of "Error: Could not find or load main class" - and you spend a couple hours trying to troubleshoot it, whether your executing a shell script that starts your java app or starting it from systemd itself - and you know your classpath is 100% correct, e.g. manually running the shell script works as well as running what you have in systemd execstart. Be sure you're running things as the correct user! In my case, I had tried different users, after quite a while of troubleshooting - i finally had a hunch, put root as the user - voila, the app started correctly. After determining it was a wrong user issue, I chown -R user:user the folder and subfolders and the app ran correctly as the specified user and group so no longer needed to run it as root (bad security).

您还可以使用监控器，这是一个非常方便的守护进程，可以用来轻松地控制服务。这些服务是由简单的配置文件定义的，这些配置文件定义了在哪个目录下哪个用户执行什么，等等，有无数的选项。supervisor ord的语法非常简单，所以它是编写SysV初始化脚本的一个很好的替代方案。

这里有一个简单的监督配置文件，用于您试图运行/控制的程序。(把这个放到/etc/supervisor/conf.d/yourapp.conf)

/etc/supervisor/conf.d/yourapp.conf

[program:yourapp]
command=/usr/bin/java -jar /path/to/application.jar
user=usertorun
autostart=true
autorestart=true
startsecs=10
startretries=3
stdout_logfile=/var/log/yourapp-stdout.log
stderr_logfile=/var/log/yourapp-stderr.log

要控制应用程序，你需要执行监控器ctl，它会提示你启动、停止和状态你的应用程序。

CLI

# sudo supervisorctl
yourapp             RUNNING   pid 123123, uptime 1 day, 15:00:00
supervisor> stop yourapp
supervisor> start yourapp

如果监控器守护进程已经在运行，并且您已经为您的服务添加了配置，而没有重新启动守护进程，那么您可以在监控器shell中简单地执行一个重读和更新命令。

这确实为您提供了使用SysV Init脚本所具有的所有灵活性，而且易于使用和控制。看一下文档。

创建一个名为your-app的脚本。服务(rest-app.service)。 我们应该把这个脚本放在/etc/systemd/system目录下。 下面是脚本的示例内容

[Unit]
Description=Spring Boot REST Application
After=syslog.target

[Service]
User=javadevjournal
ExecStart=/var/rest-app/restdemo.jar
SuccessExitStatus=200

[Install]
WantedBy=multi-user.target

下一个:

 service rest-app start

参考文献

在这里输入链接描述

I don't know of a "standard" shrink-wrapped way to do that with a Java app, but it's definitely a good idea (you want to benefit from the keep-alive and monitoring capabilities of the operating system if they are there). It's on the roadmap to provide something from the Spring Boot tool support (maven and gradle), but for now you are probably going to have to roll your own. The best solution I know of right now is Foreman, which has a declarative approach and one line commands for packaging init scripts for various standard OS formats (monit, sys V, upstart etc.). There is also evidence of people having set stuff up with gradle (e.g. here).

在systemd单元文件中，您可以通过目录或EnvironmentFile设置环境变量。我建议这样做，因为这似乎是最小的摩擦。

示例单元文件

$ cat /etc/systemd/system/hello-world.service
[Unit]
Description=Hello World Service
After=systend-user-sessions.service

[Service]
EnvironmentFile=/etc/sysconfig/hello-world
Type=simple
ExecStart=/usr/bin/java ... hello-world.jar

然后在/etc/sysconfig/hello-world下设置一个文件，其中包含Spring Boot变量的大写名称。例如，一个名为server的变量。port将遵循SERVER_PORT形式作为环境变量:

$ cat /etc/sysconfig/hello-world
SERVER_PORT=8081

这里所利用的机制是Spring Boot应用程序将获取属性列表，然后转换它们，将所有内容都改为大写，并将点替换为下划线。一旦Spring Boot应用程序完成了这个过程，它就会寻找匹配的环境变量，并相应地使用找到的环境变量。

在这篇题为:如何通过环境变量在其名称中设置带有下划线的Spring Boot属性的SO Q&A中强调了更多细节?

参考文献

第四部分。弹簧引导功能- 24。外部化配置