quartz-scheduler – 如何使用Quartz XML插件声明一次性简单触发器？

2019年8月3日 196次阅读

我正在尝试建立一个工作,使其在我的调度程序第一次启动时运行,然后在每天的午夜运行一次.这是我的工作和触发器的
XML.每天一次触发工作,但一次性触发不起作用.

    <job>
      <name>MyJob</name>
      <group>MyJobGroup</group>
      <job-type>MyScheduledJob, MyJobAssembly</job-type>
      <description>My job, yo</description>
      <durable>true</durable>
      <recover>false</recover>
    </job>
    <trigger>
      <cron>
        <name>MyTrigger</name>
        <group>MyTriggerGroup</group>
        <job-name>MyJob</job-name>
        <job-group>MyJobGroup</job-group>
        <cron-expression>0 0 0 1/1 * ? *</cron-expression>
      </cron>
        <simple>
            <name>MyOneTimeTrigger</name>
            <group>MyTriggerGroup</group>
            <description>Run once at startup, G</description>
            <misfire-instruction>SmartPolicy</misfire-instruction>
            <volatile>false</volatile>
            <job-name>MyJob</job-name>
            <job-group>MyJobGroup</job-group>
            <repeat-count>0</repeat-count>
            <repeat-interval>0</repeat-interval> 
      </simple>
    </trigger>

最佳答案 通过不起作用我想你的意思是触发器永远不会触发.也许是因为它没有延迟它会失败？如何添加< start-time-seconds-in-future />在未来的某个时间？

<simple>
        <name>MyOneTimeTrigger</name>
        <group>MyTriggerGroup</group>
        <description>Run once at startup, G</description>
        <misfire-instruction>SmartPolicy</misfire-instruction>
        <volatile>false</volatile>
        <job-name>MyJob</job-name>
        <job-group>MyJobGroup</job-group>
        <start-time-seconds-in-future>60</start-time-seconds-in-future>
        <repeat-count>0</repeat-count>
        <repeat-interval>0</repeat-interval> 
</simple>

如果它适合您,您必须知道您可能认为是错误或功能的事情：如果您重新启动服务器,因为触发器已经被触发并被删除,它将再次触发.见：Quartz XML plugin reschedules fired triggers after restart.

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