Spring + Quartz重新安排或更新触发器？ [英] Spring + Quartz resheduling or updating the trigger?

问题描述

我正在使用 Spring 3.2 和 Quartz 2.2 。

我的目标类和方法，

public class Producer {

    public void executeListener() {
        System.out.println(" Test Method . ");
    }
}

在我的春天 applicationContext。 xml ，

<bean id="producer" class="com.service.Producer" />

<bean id="jobDetail" class="org.springframework.scheduling.quartz.MethodInvokingJobDetailFactoryBean">
  <property name="targetObject" ref="producer" />
  <property name="targetMethod" value="executeListener" />
  <property name="concurrent" value="false" />
</bean>

<bean id="simpleTrigger" class="org.springframework.scheduling.quartz.SimpleTriggerFactoryBean">
    <property name="jobDetail" ref="jobDetail" />
    <!-- 10 seconds -->
    <property name="startDelay" value="10000" />
    <!-- repeat every 5 seconds -->
    <property name="repeatInterval" value="5000" />
</bean>


<bean id="mySheduler" class="org.springframework.scheduling.quartz.SchedulerFactoryBean">
    <property name="triggers">
    <list>
        <!-- <ref bean="cronTrigger" /> -->
        <ref bean="simpleTrigger" />
    </list>
    </property>
</bean>

以上代码正常运行， executeListener（） nethod连续执行，间隔为 5秒，启动延迟为 10秒。

The above code is working and the executeListener() nethod is executing continously with the interval of 5 seconds and the starting delay will be 10 sec.

我已尝试在 spring controller 中更新触发器，

I have tried in spring controller to update the trigger as ,

@Controller
public class SpringController {

    @Autowired
    org.springframework.scheduling.quartz.SchedulerFactoryBean myScheduler;

    @Autowired
    org.springframework.scheduling.quartz.SimpleTriggerFactoryBean oldTrigger;

    @RequestMapping(value="/reSheduleTrigger.html",method=RequestMethod.POST)
    public @ResponseBody String reSheduleTrigger(@RequestParam(value="triggerInSec") String triggerInSec ){

        System.out.println("----------------- Changing the repeat interval request -----------");

        // obtain a builder that would produce the trigger
        TriggerBuilder<SimpleTrigger> tb = oldTrigger.getObject().getTriggerBuilder();   

        Trigger newTrigger = tb.withSchedule(simpleSchedule().withIntervalInSeconds(triggerInSec).build());

        myScheduler.getObject().rescheduleJob(oldTrigger.getObject().getKey(),newTrigger  );

        return "success";
    }
}

但我收到编译时错误触发newTrigger = tb.withSchedule（simpleSchedule（）。withIntervalInSeconds（triggerInSec）.build（））; 。

我需要：

我想重新计算执行的工作executeListener（）动态 2秒区间的方法。

I want to reshedule the job to execute the executeListener() method with the 2 sec interval dynamically.

谢谢。

推荐答案

请尝试以下方法之一：

Trigger newTrigger = tb.withSchedule(SimpleScheduleBuilder.simpleSchedule().withIntervalInSeconds(2)).build();

或

Trigger newTrigger = tb.withSchedule(SimpleScheduleBuilder.repeatSecondlyForever(2)).build();

两者在语法上都有效。看起来你的问题是触发newTrigger = tb.withSchedule（simpleSchedule（）。withIntervalInSeconds（triggerInSec）.build（））; 是由不正确的包围引起的。即build（）之后不应该有一个额外的括号，因为它返回一个触发器。

Both are syntactically valid. It looks like your issue with Trigger newTrigger = tb.withSchedule(simpleSchedule().withIntervalInSeconds(triggerInSec).build()); is caused by incorrect bracketing. ie build() should not have an extra bracket after it, as it returns a trigger.

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