在Linux中使用systemd启动脚本 [英] Startup script with systemd in Linux
本文介绍了在Linux中使用systemd启动脚本的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我可以在下面执行此启动服务吗,一旦运行就没有错误显示,但是下面的服务器脚本无法运行!
Can I do This start up service below, there are no errors showing once run, but the server script below does not run!
ln /lib/systemd/aquarium.service aquarium.service
systemctl daemon-reload
systemctl enable aquarium.service
systemctl start aquarium.service
谢谢
aquarium.service:
[Unit]
Description=Start aquarium server
[Service]
WorkingDirectory=/home/root/python/code/aquarium/
ExecStart=/bin/bash server.* start
KillMode=process
[Install]
WantedBy=multi-user.target
这是server.sh脚本
#!/bin/bash
PID=""
function get_pid {
PID=`pidof python ./udpthread.py`
}
function stop {
get_pid
if [ -z $PID ]; then
echo "server is not running."
exit 1
else
echo -n "Stopping server.."
kill -9 $PID
sleep 1
echo ".. Done."
fi
}
function start {
get_pid
if [ -z $PID ]; then
echo "Starting server.."
./udpthread.py &
get_pid
echo "Done. PID=$PID"
else
echo "server is already running, PID=$PID"
fi
}
function restart {
echo "Restarting server.."
get_pid
if [ -z $PID ]; then
start
else
stop
sleep 5
start
fi
}
function status {
get_pid
if [ -z $PID ]; then
echo "Server is not running."
exit 1
else
echo "Server is running, PID=$PID"
fi
}
case "$1" in
start)
start
;;
stop)
stop
;;
restart)
restart
;;
status)
status
;;
*)
echo "Usage: $0 {start|stop|restart|status}"
esac
推荐答案
尝试使用"Type = forking"并使用完整的文件名.
Try using "Type=forking" and use complete filename.
[Unit]
Description=Start aquarium server
[Service]
WorkingDirectory=/home/root/python/code/aquarium/
Type=forking
ExecStart=/bin/bash server.sh start
KillMode=process
[Install]
WantedBy=multi-user.target
如果不起作用,请发布此命令的输出:
if it not work, post output of this command:
# journalctl -u aquarium.service
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