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Monit：ping 网站 URL 然后重新启动应用程序服务器

tag-icon • tag-icon linux ubuntu monit
Monit：ping 网站 URL 然后重新启动应用程序服务器

我正在使用此配置来监控 apache2（ubuntu）：

/etc/monit/conf.d/apache2：

check process apache with pidfile /var/run/apache2/apache2.pid

    start program = "/etc/init.d/apache2 start" with timeout 60 seconds
    stop program  = "/etc/init.d/apache2 stop"

    alert [email protected] with mail-format {
            from: [email protected]
            subject: monit alert: $SERVICE $EVENT $DATE
            message: $DESCRIPTION
    }

    if failed port 80 protocol HTTP
            request /
            with timeout 7 seconds
            then restart

访问“request /”将进入 ubuntu-apache 默认页面：如果您可以访问它，则表明 Apache 正在运行。这样就没问题了。

现在我还需要监控我的应用服务器。所以我有第二个配置文件：

check process lucee with pidfile /opt/lucee/tomcat/work/tomcat.pid

        start program = "/etc/init.d/lucee_ctl start" with timeout 60 seconds
        stop program  = "/etc/init.d/lucee_ctl stop"

我不知道如何告诉 monit 请求完全合格的 URI，我需要类似

if failed port 80 protocol HTTP
            request http://www.xxxxxx123123.com/
            with timeout 7 seconds
            then restart

但这说明

/etc/monit/conf.d/lucee:13：错误：语法错误'

重新加载。有什么想法吗？

答案1

根据样本

  if failed host localhost port 80 with protocol http and request "/server-status" with timeout 25 seconds for 4 times within 5 cycles then restart

尝试改为

  if failed host www.xxxxxx123123.com port 80 with protocol http 
      and request "/" 
      with timeout 7 seconds 
      then restart

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