求助 quartz启动与暂停问题

[隋春羽]

近日由于项目需要用到quartz组件实现调度管理功能，想要实现调度的启动和暂停以及恢复功能，但是暂停遇到问题
暂停后重新启动，会连续多次调用job中的execute方法。如果当前工作的处理时间过长必然会导致问题。代码如下，急求帮助

import java.util.Date;

import org.quartz.JobDetail;
import org.quartz.Scheduler;
import org.quartz.SchedulerException;
import org.quartz.SchedulerFactory;
import org.quartz.SimpleTrigger;
import org.quartz.impl.StdSchedulerFactory;

public class TestTwo {

    /**
     * @param args
     */
    public static void main(String[] args) {
        
        try {
            SchedulerFactory sf = new StdSchedulerFactory(); 
            Scheduler sched = sf.getScheduler();
            
             JobDetail job = new JobDetail();
             job.setJobClass(HelloJob.class);
             job.setGroup("jobgourp");
             job.setName("jobname");

             SimpleTrigger striger=new SimpleTrigger();
             striger.setName("strigername");
             striger.setStartTime(new Date());
             striger.setRepeatInterval(1000);
             striger.setRepeatCount(-1);
             
             striger.setJobName("jobname");
             striger.setJobGroup("jobgourp");
             
             sched.scheduleJob(job, striger);
             sched.start();
             //-------------------------------------------------------
             Thread.sleep(4000);
             System.out.println("暂停");
             
             sched.pauseJobGroup("jobgourp");
             sched.pauseTriggerGroup("jobgourp");
             Thread.sleep(5000);
             System.out.println("暂停结束");
             
             sched.resumeJobGroup("jobgourp");
             sched.resumeTriggerGroup("jobgourp");
             sched.start();
             
             Thread.sleep(4800);
             sched.shutdown(true);
        } catch (SchedulerException e) {
            e.printStackTrace();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } 
    }

}

job 类代码如下

package com.scy.quartz.test;

import java.text.SimpleDateFormat;
import java.util.Date;

import org.quartz.Job;
import org.quartz.JobExecutionContext;
import org.quartz.JobExecutionException;

public class HelloJob implements Job {

    @Override
    public void execute(JobExecutionContext arg0) throws JobExecutionException {
            System.out.println("HelloJob 执行了"+new SimpleDateFormat("yyyy-MM-dd hh:mm:ss").format(new Date()));
        
    }

}

HelloJob 执行了2012-08-11 12:57:16
HelloJob 执行了2012-08-11 12:57:17
HelloJob 执行了2012-08-11 12:57:18
HelloJob 执行了2012-08-11 12:57:19
暂停
暂停结束
----------????-----------
HelloJob 执行了2012-08-11 12:57:25
HelloJob 执行了2012-08-11 12:57:25
HelloJob 执行了2012-08-11 12:57:25
HelloJob 执行了2012-08-11 12:57:25
HelloJob 执行了2012-08-11 12:57:25
HelloJob 执行了2012-08-11 12:57:25
----------????-----------
HelloJob 执行了2012-08-11 12:57:26
HelloJob 执行了2012-08-11 12:57:27
HelloJob 执行了2012-08-11 12:57:28
HelloJob 执行了2012-08-11 12:57:29

[MiceRice]

resumeJobGroup的API解释说了（错过的触发，将在恢复时重新执行）：
If any of the Job'sTrigger s missed one or more fire-times, then the Trigger's misfire instruction will be applied.

所以我想你应该用standby()会更合适（但这个影响是全局性的）：
When start() is called (to bring the scheduler out of stand-by mode), trigger misfire instructions will NOT be applied during the execution of the start() method - any misfires will be detected immediately afterward (by the JobStore's normal process).

或者用unscheduleJob()将这个任务撤掉。


顺带说一句：在你这个例子里面，pauseJobGroup() 跟 pauseTriggerGroup() 重复，因为pauseJobGroup实际上也是暂停它的triggers：
Pause all of the JobDetails in the given group - by pausing all of their Triggers.