2017-07-10 58 views
2

我试图运行花朵守护进程。读作我的flower.service文件如下:如何在RHEL中启动芹菜花作为守护进程?

[Unit] 

    Description=Flower Service 
    After=network.target 

[Service] 

    Type=forking 
    User=maas 
    Group=maas 
    PermissionsStartOnly=true 
    ExecStart=/bin/flower --broker=amqp://oser000300// 

[Install] 

    WantedBy=multi-user.target 

但是,当我启动服务,它给错误。

//systemctl status flower.service 

* flower.service - Flower Service 
    Loaded: loaded (/etc/systemd/system/flower.service; enabled; vendor preset: disabled) 
    Active: failed (Result: timeout) since Mon 2017-07-10 20:25:59 UTC; 4min 38s ago 
    Process: 49255 ExecStart=/bin/flower --broker=amqp://oser000300// (code=exited, status=0/SUCCESS) 


Connected to amqp://guest:**@oser000300:5672// 
flower.service start operation timed out. Terminating. 
SIGTERM detected, shutting down 
Failed to start Flower Service. 
Unit flower.service entered failed state. 
flower.service failed. 
+1

花是芹菜监控工具(作品使用Redis的),你运行芹菜? – jmugz3

+0

是的,芹菜正在运行,并能够正常运行花,但无法启动作为服务。 – Datta

+0

@jww我认为这是一个编程问题,我的问题在于编写shell脚本或其他任何形式的服务文件。 – Datta

回答

0

我在启动服务时遇到同样的超时问题。

这些参数已经做的伎俩(我已经芹菜服务与派生类型的运行):

Type=simple 
Restart=on-failure