在群集环境中创建Quartz触发器

Question

相关：Quartz群集-服务器启动时触发器重复

private void createOrUpdateJob(JobKey jobKey, Class<? extends org.quartz.Job> clazz, Trigger trigger) throws SchedulerException {
    JobBuilder jobBuilder = JobBuilder.newJob(clazz).withIdentity(jobKey);
    if (!scheduler.checkExists(jobKey)) {
        // if the job doesn't already exist, we can create it, along with its trigger. this prevents us
        // from creating multiple instances of the same job when running in a clustered environment
        scheduler.scheduleJob(jobBuilder.build(), trigger);
        log.error("SCHEDULED JOB WITH KEY " + jobKey.toString());
    } else {
        // if the job has exactly one trigger, we can just reschedule it, which allows us to update the schedule for
        // that trigger.
        List<? extends Trigger> triggers = scheduler.getTriggersOfJob(jobKey);
        if (triggers.size() == 1) {
            scheduler.rescheduleJob(triggers.get(0).getKey(), trigger);
            return;
        }

        // if for some reason the job has multiple triggers, it's easiest to just delete and re-create the job,
        // since we want to enforce a one-to-one relationship between jobs and triggers
        scheduler.deleteJob(jobKey);
        scheduler.scheduleJob(jobBuilder.build(), trigger);
    }
}

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这是很好的，但我担心的是，当两个实例完全同时启动时，可能会出现竞争情况。因为在这段代码周围没有全局锁，集群中的所有节点都将尊重它，如果两个实例同时联机，我可能会得到重复的作业或触发器，这就违背了这段代码的要点。

在集群环境中是否有自动定义Quartz作业和触发器的最佳实践？还是我需要自己设置锁？

Answer 1

我不确定是否有更好的方法在石英中做到这一点。但如果您已经在使用Redis或Memcache，我建议让所有实例对一个众所周知的键执行原子增量。如果粘贴的代码每小时每个群集只运行一个作业,则可以执行以下操作：

long timestamp = System.currentTimeMillis() / 1000 / 60 / 60;
String key = String.format("%s_%d", jobId, timestamp);

// this will only be true for one instance in the cluster per (job, timestamp) tuple
bool shouldExecute = redis.incr(key) == 1

if (shouldExecute) {
  // run the mutually exclusive code
}

时间戳为您提供了一个移动窗口,在该窗口中,作业正在竞争执行该作业。