Question

我想要做的是：按一个按钮改变简单值。如果值保持3秒不变，我想执行我的方法。所以我需要一个单独的线程，它会在每次点击按钮时启动（并杀死所有其他线程）。我知道

Executor executor = Executors.newSingleThreadScheduledExecutor();

但它不会有延迟。这样做的最佳做法是什么？

Answer 1

您可以简单地使用旧的Thread类或Runnable接口：单击按钮后立即启动线程，执行Thread.sleep(3000)作为第一条指令，然后检查值是否为没有改变。在开始线程之前，检查另一个线程是否已经存在，如果有，请在其上调用interrupt()并处理InterruptedException以正常终止它。

如果您想使用较新的并发API，请注意Executors.newSingleThreadScheduledExecutor()返回ScheduledExecutorService（不是简单的Executor）并且它允许启动具有延迟的线程，顾名思义。查看Javadoc。

Answer 2

您可以使用简单的Thread启停/中断功能，也可以使用ScheduledExecutorService之类的API。如何使用ScheduledExecutorService的示例位于

之下

//Start as many threads as you want upon button click... (since you said i want to kill all other threads, so i assume you want to start some threads soon upon button click)
            ScheduledExecutorService service = Executors.newScheduledThreadPool(3);
            service.schedule(new Thread(), 0, TimeUnit.SECONDS);
            service.schedule(new Thread(), 0, TimeUnit.SECONDS);
            service.schedule(new Thread(), 0, TimeUnit.SECONDS);

            //Start single thread after 3 seconds.
            ScheduledExecutorService serviceCheck = Executors.newScheduledThreadPool(3);
            serviceCheck.schedule(new Thread(), 3, TimeUnit.SECONDS);
            //In this thread created using "serviceCheck", check with value has not changed then use `service.shutdownNow();` to shutdown all previously started threads and then do whatever you want.

单线程有延迟启动

2 个答案: