Question

我想安排作业以2秒的倍数运行，即2,4,8,16,32秒。第二次火灾应在第一次火灾完成两秒后发生，第三次火灾应在第二次火灾完成4秒后发生，依此类推。下一场火灾是基于我们从之前的火力中获得的状态，根据该状态判断我们是否需要触发下一次火力。有人可以告诉我如何使用石英调度程序来实现这一目标？

如果我使用SimpleTrigger.withIntervalInSeconds（2）它会在每2秒后运行一个作业，因为我希望时间间隔应该在每次发射时增加2的倍数。

Answer 1

也许您可能忘记尝试设置单个触发器，但使用多个触发器。我的Java在这方面并不好，所以这是伪代码：

delay = 2
repeat
 TriggerOnceAfter(delay)
 delay <- delay * 2
 WaitUntilTriggered()
until (finished)

我不确定如何实施WaitUntilTriggered()方法;你需要在WaitUntilTriggered()的触发代码中添加一个信令标志来查看。

这将延迟2,4,8，......

Answer 2

这是一个简化的实现，它将按照请求的时间表调用Runnable：

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.TimeUnit;

public class Tasker {

    private int numberOfRuns; //how many times job executed 
    private int timeBetweenRuns;//seconds

    Tasker(int numberOfRuns){

        this.numberOfRuns = numberOfRuns;
        timeBetweenRuns = 2;
        execute();
    }


    private void execute() {

        for (int counter = 0; counter < numberOfRuns ; counter++) {
            CountDownLatch latch = new CountDownLatch(1);
            Job job = new Job(latch, timeBetweenRuns);
            job.run();

            try {
                latch.await();
                TimeUnit.SECONDS.sleep(timeBetweenRuns);
            } catch (InterruptedException ex) {
                ex.printStackTrace();
            }
            timeBetweenRuns *=2;
        }
    }

    public static void main(String[] args){

        new Tasker(5);
    }
}

class Job implements Runnable {

    private int seconds;
    private CountDownLatch latch ;

    Job(CountDownLatch latch , int seconds){
        this.latch = latch;
        this.seconds = seconds;
    }

    @Override
    public void run() {

        System.out.println("Job runs "+ seconds +" after previous one");
        latch.countDown();
    }
}

安排工作以2秒的倍数运行，即2,4,8,16,32秒

2 个答案: