Question

如果工作抛出异常，我已经搜索了如何在一段时间后重新开始工作的答案。我看不出这样做的简单方法。

如果我像这样设置我的触发器：

JobDetail job = new JobDetail("Download catalog", null, typeof(MyJob));
job .Durable = true;
Trigger trigger= TriggerUtils.MakeDailyTrigger(12, 0);
trigger.StartTimeUtc = DateTime.UtcNow;
trigger.Name = "trigger name";
scheduler.ScheduleJob(job , trigger);

MyJob看起来像这样：

public class MyJob : IJob
{
    public void Execute(JobExecutionContext context)
    {
        var service = new service();


        try
        {
            service.Download();
        }
        catch (Exception)
        {
            throw;
        }

    }
}

如果service.Download（）调用抛出某种异常，如何在15分钟后触发重新启动/重新触发？

Answer 1

实际上，没有必要像LeftyX那样创建一个新的JobDetail。您可以只安排从当前上下文连接到JobDetail的新触发器。

public void Execute(JobExecutionContext context) {
    try {
        // code
    } catch (Exception ex) {
        SimpleTriggerImpl retryTrigger = new SimpleTriggerImpl(Guid.NewGuid().ToString());      
        retryTrigger.Description = "RetryTrigger";
        retryTrigger.RepeatCount = 0;
        retryTrigger.JobKey = context.JobDetail.Key;   // connect trigger with current job      
        retryTrigger.StartTimeUtc = DateBuilder.NextGivenSecondDate(DateTime.Now, 30);  // Execute after 30 seconds from now
        context.Scheduler.ScheduleJob(retryTrigger);   // schedule the trigger

        JobExecutionException jex = new JobExecutionException(ex, false);
        throw jex;
    }
}

这比创建新的JobDetail更不容易出错。希望有所帮助。

Answer 2

我认为正确的答案是使用JobListener重试作业，如下所述：http://thecodesaysitall.blogspot.cz/2012/03/quartz-candy-part-1.html。

在此解决方案中将重试逻辑与Job本身分开，因此可以重复使用。

如果您按照此处其他回复中的建议在作业中实施重试逻辑，则必须在每个作业中再次实施。

编辑：根据Ramanpreet Singh的说明，可以找到更好的解决方案here： https://blog.harveydelaney.com/quartz-job-exception-retrying/

Answer 3

我认为你唯一的选择是捕获错误并告诉Quartz.net立即重新启用：

public class MyJob : IJob
{
    public void Execute(JobExecutionContext context)
    {
        var service = new service();

        try
        {
            service.Download();
        }
        catch (Exception ex)
        {
              JobExecutionException qe = new JobExecutionException(ex);
              qe.RefireImmediately = true;  // this job will refire immediately
              throw qe;  
        }
    }
}

您可以找到一些信息here和here。

<强>更新：

我做了一些测试，似乎你可以在执行的工作中安排一个新的触发器你可以尝试这样的事情：

public class MyJob : IJob
{
    public void Execute(JobExecutionContext context)
    {
        var service = new service();

        try
        {
            service.Download();
        }
        catch (Exception ex)
        {
            JobExecutionException qe = new JobExecutionException(ex);
            // qe.RefireImmediately = true;  // this job will refire immediately
            // throw qe;  
            OnErrorScheduleJob(context);

        }
    }

    private void OnErrorScheduleJob(JobExecutionContext context)
    {
        var jobOnError = context.Scheduler.GetJobDetail("ONERRORJOB", "ERROR");
        if (jobOnError == null)
        {
        JobDetail job = new JobDetail("ONERRORJOB", "ERROR", typeof(MyJob));
        job.Durable = false;
        job.Volatile = false;
        job.RequestsRecovery = false;

        SimpleTrigger trigger = new SimpleTrigger("ONERRORTRIGGER",
                        "ERROR",
                        DateTime.UtcNow.AddMinutes(15),
                        null,
                        1,
                        TimeSpan.FromMinutes(100));

        context.Scheduler.ScheduleJob(job, trigger);     
        }
    }
}

Answer 4

// don't forget to use @PersistJobDataAfterExecution without it, the jobExecutionContext will reset the value of count.     

SimpleTriggerImpl retryTrigger = new SimpleTriggerImpl();
    retryTrigger.setName("jobname");
    retryTrigger.setRepeatCount(0);
    retryTrigger.setJobKey(jobExecutionContext.getJobDetail().getKey());
    final Calendar cal = getCalendarInstance();
    cal.add(Calendar.MINUTE, 1); //retry after one minute
    retryTrigger.setStartTime(cal.getTime());
    try {
        jobExecutionContext.getScheduler().scheduleJob(retryTrigger);   // schedule the trigger
    } catch (SchedulerException ex) {
        logger.error("something went wrong", ex); 
    }
    JobExecutionException e2 = new JobExecutionException("retrying...");
    e2.refireImmediately();
    throw e2;

如果作业失败，则在15分钟后激活quartz.net触发器

4 个答案: