Spring + Quartz重新安排或更新触发器？

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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.