如何在Linux系统中将Spring Boot应用程序打包为可执行jar as a Service ?这是推荐的方法吗,还是应该将这个应用程序转换为war并将其安装到Tomcat中?
目前,我可以从屏幕会话运行Spring引导应用程序,这很好,但需要在服务器重新启动后手动启动。
我正在寻找的是一般的建议/方向或样本init。D脚本,如果我的方法与可执行jar是适当的。
如何在Linux系统中将Spring Boot应用程序打包为可执行jar as a Service ?这是推荐的方法吗,还是应该将这个应用程序转换为war并将其安装到Tomcat中?
目前,我可以从屏幕会话运行Spring引导应用程序,这很好,但需要在服务器重新启动后手动启动。
我正在寻找的是一般的建议/方向或样本init。D脚本,如果我的方法与可执行jar是适当的。
当前回答
下面是一个脚本,它将可执行jar部署为systemd服务。
它为服务和.service文件创建一个用户,并将jar文件放在/var下,并对特权进行一些基本的锁定。
#!/bin/bash
# Argument: The jar file to deploy
APPSRCPATH=$1
# Argument: application name, no spaces please, used as folder name under /var
APPNAME=$2
# Argument: the user to use when running the application, may exist, created if not exists
APPUSER=$3
# Help text
USAGE="
Usage: sudo $0 <jar-file> <app-name> <runtime-user>
If an app with the name <app-name> already exist, it is stopped and deleted.
If the <runtime-user> does not already exist, it is created.
"
# Check that we are root
if [ ! "root" = "$(whoami)" ]; then
echo "Must be root. Please use e.g. sudo"
echo "$USAGE"
exit
fi
# Check arguments
if [ "$#" -ne 3 -o ${#APPSRCPATH} = 0 -o ${#APPNAME} = 0 -o ${#APPUSER} = 0 ]; then
echo "Incorrect number of parameters."
echo "$USAGE"
exit
fi
if [ ! -f $APPSRCPATH ]; then
echo "Can't find jar file $APPSRCPATH"
echo "$USAGE"
exit
fi
# Infered values
APPFILENAME=$(basename $APPSRCPATH)
APPFOLDER=/var/javaapps/$APPNAME
APPDESTPATH=$APPFOLDER/$APPFILENAME
# Stop the service if it already exist and is running
systemctl stop $APPNAME >/dev/null 2>&1
# Create the app folder, deleting any previous content
rm -fr $APPFOLDER
mkdir -p $APPFOLDER
# Create the user if it does not exist
if id "$APPUSER" >/dev/null 2>&1; then
echo "Using existing user $APPUSER"
else
adduser --disabled-password --gecos "" $APPUSER
echo "Created user $APPUSER"
fi
# Place app in app folder, setting owner and rights
cp $APPSRCPATH $APPDESTPATH
chown $APPUSER $APPDESTPATH
chmod 500 $APPDESTPATH
echo "Added or updated the $APPDESTPATH file"
# Create the .service file used by systemd
echo "
[Unit]
Description=$APPNAME
After=syslog.target
[Service]
User=$APPUSER
ExecStart=/usr/bin/java -jar $APPDESTPATH
SuccessExitStatus=143
[Install]
WantedBy=multi-user.target
" > /etc/systemd/system/$APPNAME.service
echo "Created the /etc/systemd/system/$APPNAME.service file"
# Reload the daemon
systemctl daemon-reload
# Start the deployed app
systemctl start $APPNAME
systemctl status $APPNAME
例子:
其他回答
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
我试图使springboot应用程序呈现为“init”。D”风格的shell脚本与压缩Java应用程序钉在最后
通过符号链接这些脚本从/etc/init.D /spring-app到/opt/spring-app.jar,并chmod jar使其可执行。D /spring-app启动/etc/init。D /spring-app stop”和其他可能的状态工作
假设是init。来自springboot的d风格脚本看起来他们有必要的魔法字符串(像# Default-Start: 2 3 4 5) chkconfig将能够将其作为“服务”添加。
但是我想让它和systemd一起工作
为了做到这一点,我尝试了上面其他答案中的许多食谱,但在Centos 7.2和Springboot 1.3上,它们都不适合我。大多数情况下,它们会启动服务,但无法跟踪pid
最后,我发现下面的方法对我有用,当/etc/init.D链接也到位了。一个类似于下面的文件应该安装为/usr/lib/systemd/system/spring-app.service
[Unit]
Description=My loverly application
After=syslog.target
[Service]
Type=forking
PIDFile=/var/run/spring-app/spring-app.pid
ExecStart=/etc/init.d/spring-app start
SuccessExitStatus=143
[Install]
WantedBy=multi-user.target
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)
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).
在这个问题中,@PbxMan的回答可以让你开始:
在Linux上运行Java应用程序作为服务
编辑:
还有另一种不太好的方式在重启时启动进程,使用cron:
@reboot user-to-run-under /usr/bin/java -jar /path/to/application.jar
这是可行的,但不能为应用程序提供良好的启动/停止界面。你仍然可以简单地杀死它……