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

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:

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

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

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).