您正在使用Maven吗?那么你应该试试AppAssembler插件:

Application Assembler Plugin是一个Maven插件，用于生成启动java应用程序的脚本. ...所有工件(依赖项+项目中的工件)都被添加到生成的bin脚本中的类路径中。 支持平台: unix变体 Windows NT(不支持Windows 9x) Java服务包装器(JSW)

参见:http://mojo.codehaus.org/appassembler/appassembler-maven-plugin/index.html

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).

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).

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

下面是在Linux中将Java应用程序作为系统服务安装的最简单方法。

让我们假设你正在使用systemd(现在任何现代发行版都是这样):

首先，在/etc/systemd/system目录下创建一个服务文件，例如javaservice。服务内容如下:

[Unit]
Description=Java Service

[Service]
User=nobody
# The configuration file application.properties should be here:
WorkingDirectory=/data 
ExecStart=/usr/bin/java -Xmx256m -jar application.jar --server.port=8081
SuccessExitStatus=143
TimeoutStopSec=10
Restart=on-failure
RestartSec=5

[Install]
WantedBy=multi-user.target

其次，通知systemd新的服务文件:

systemctl daemon-reload

并启用它，以便它在引导时运行:

systemctl enable javaservice.service

最后，您可以使用以下命令来启动/停止您的新服务:

systemctl start javaservice
systemctl stop javaservice
systemctl restart javaservice
systemctl status javaservice

如果使用systemd，这是将Java应用程序设置为系统服务的最非侵入性和最干净的方法。

我特别喜欢这个解决方案的地方在于，您不需要安装和配置任何其他软件。所提供的systemd为您完成所有工作，您的服务就像任何其他系统服务一样。我现在在生产环境中使用了一段时间，在各种发行版上，它就像你期望的那样工作。

另一个优点是，通过使用/usr/bin/java，您可以轻松添加jvm参数，如-Xmx256m。

请阅读官方Spring Boot文档中的systemd部分: http://docs.spring.io/spring-boot/docs/current/reference/html/deployment-install.html

如果你想使用Spring Boot 1.2.5和Spring Boot Maven Plugin 1.3.0。M2，这是我们的解

<parent>
    <groupId>org.springframework.boot</groupId>
    <artifactId>spring-boot-starter-parent</artifactId>
    <version>1.2.5.RELEASE</version>
</parent>

<build>
    <plugins>
        <plugin>
            <groupId>org.springframework.boot</groupId>
            <artifactId>spring-boot-maven-plugin</artifactId>
            <version>1.3.0.M2</version>
            <configuration>
                <executable>true</executable>
            </configuration>
        </plugin>
    </plugins>
</build>

<pluginRepositories>
    <pluginRepository>
        <id>spring-libs-milestones</id>
        <url>http://repo.spring.io/libs-milestone</url>
    </pluginRepository> 
</pluginRepositories>

然后像往常一样编译:mvn清洁包，做一个符号链接ln -s /…/myapp.jar /etc/init.chmod +x /etc/init. D /myapp，使其可执行启动myapp服务(Ubuntu服务器)