在执行程序服务延迟后安排任务

Question

我希望在延迟3秒后执行任务，而我的一项任务需要2秒才能完成。

我得到的输出显示5秒的间隔

注意：Student类实现了Callable接口， 我有以下查询

1. 为什么会有5秒的延迟。如何才能使3秒的延迟为何在第二次执行中显示线程1，所以应该是线程2

我得到的输出是

The time is : Sat Nov 26 15:08:02 IST 2016

Doing a task during : prerna - Time - Sat Nov 26 15:08:06 IST 2016
pool-1-thread-1 Helloprerna
Doing a task during : abc - Time - Sat Nov 26 15:08:11 IST 2016
pool-1-thread-1 Helloabc
Doing a task during : def - Time - Sat Nov 26 15:08:16 IST 2016
pool-1-thread-2 Hellodef
Doing a task during : xyz - Time - Sat Nov 26 15:08:21 IST 2016
pool-1-thread-1 Helloxyz
Doing a task during : ritu - Time - Sat Nov 26 15:08:26 IST 2016
pool-1-thread-3 Helloritu
Doing a task during : babita - Time - Sat Nov 26 15:08:31 IST 2016
pool-1-thread-2 Hellobabita

代码：

private String display(String name2) {

    try {
        //  System.out.println(Thread.currentThread().getName());
        name2=Thread.currentThread().getName()+" Hello"+ name;
        System.out.println("Doing a task during : " + name + " - Time - " + new Date());
        Thread.sleep(000);
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return name2;
}


@Override
public String call() throws Exception {
    // TODO Auto-generated method stub
    if (name == "archana") {

        throw new Exception();
    }
        /*} catch (Exception e) {
            // TODO Auto-generated catch block
        //  e.printStackTrace();
        }finally{
            return "error";
        }*/

    return display(name);
}

public class ExecutorScheduleDemo {

    public static void main(String args[]) throws InterruptedException{
        ScheduledExecutorService executor= Executors.newScheduledThreadPool(5);
        ArrayList<Student> list = new ArrayList<Student>();

        list.add(new Student("prerna"));
        list.add(new Student("abc"));
        //list.add(new Student("archana"));
        list.add(new Student("def"));
        list.add(new Student("xyz"));
        list.add(new Student("ritu"));
        list.add(new Student("babita"));
        System.out.println("The time is : " + new Date());
        List<Future<String>> resultList= new  ArrayList<Future<String>>();
        for(Student s:list){
            Future<String> f=executor.schedule(s, 3, TimeUnit.SECONDS);

            try {
                System.out.println(f.get());
            } catch (ExecutionException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
    }
}

Answer 1

使用scheduleAtFixedRate(Runnable, long initialDelay, long period, TimeUnit timeunit)代替schedule(Runnable task, long delay, TimeUnit timeunit)。

scheduleAtFixedRate (Runnable, long initialDelay, long period, TimeUnit timeunit)
创建并执行一个周期性操作，该操作在给定的初始延迟后首先启用，然后在给定的时间段内启用；也就是说执行将在initialDelaythen initialDelay+period，then等之后开始initialDelay + 2 * period。如果该任务的任何执行遇到异常，则将禁止后续执行。否则，任务将仅通过取消或终止执行程序而终止。如果此任务的任何执行花费的时间超过其时间，则后续执行可能会开始得较晚，但不会同时执行。下一次执行。