我想从源代码部署一个 play 框架 Web 应用程序,并运行play start
以启动该应用程序。
我写了一个启动脚本,在服务启动时/etc/init.d/
执行,但服务启动命令没有返回。daemon play start
我想这是因为play start
正在等我输入Ctrl+ D。nohup
可以修复它,但是对于nohup
,我必须运行kill -9 xxx
才能停止应用程序,这不是我喜欢的。
从源代码运行 play 框架应用程序作为 Linux 启动服务的最佳方法是什么?
答案1
这是简单的init.d
脚本,其中:
start
:仅当应用程序尚未启动时,重新编译(如果需要)并在后台启动应用程序stop
: 杀死应用程序
仔细阅读代码中的注释。
#!/bin/sh
# /etc/init.d/playapp
# Play project directory is in /var/play/playapp/www, not directly in SDIR
SDIR="/var/play/playapp"
# The following part always gets executed.
echo "PLAYAPP Service"
# The following part carries out specific functions depending on arguments.
case "$1" in
start)
echo " * Starting PLAYAPP Service"
if [ -f ${SDIR}/www/target/universal/stage/RUNNING_PID ]
then
PID=$(cat ${SDIR}www/target/universal/stage/RUNNING_PID)
if ps -p $PID > /dev/null
then
echo " service already running ($PID)"
exit 1
fi
fi
cd ${SDIR}/www
# REPLACE "PROJECT_NAME" with your project name
if [ ! -f ${SDIR}/www/target/universal/stage/bin/PROJECT_NAME ]
then
echo " recompiling..."
# REPLACE path to your play command
/var/play-install/play/play clean compile stage
fi
echo " starting..."
nohup ./target/universal/stage/bin/PROJECT_NAME -Dhttp.port=9900 -Dconfig.file=/var/play/playapp/www/conf/application-prod.conf > application.log 2>&1&
;;
stop)
echo " * Stopping PLAYAPP Service"
if [ ! -f ${SDIR}/www/target/universal/stage/RUNNING_PID ]
then
echo " nothing to stop"
exit 1;
fi
kill -TERM $(cat ${SDIR}/www/target/universal/stage/RUNNING_PID)
;;
*)
echo "Usage: /etc/init.d/playapp {start|stop}"
exit 1
;;
esac
exit 0