Question

这将每5秒发出一次滴答声。

Observable.interval(5, TimeUnit.SECONDS, Schedulers.io())
            .subscribe(tick -> Log.d(TAG, "tick = "+tick));

要停止它，您可以使用

Schedulers.shutdown();

然后所有调度程序停止，以后无法恢复滴答。如何停止和恢复发射＆＃34;优雅＆＃34;`？

Answer 1

这是一个可能的解决方案：

class TickHandler {

    private AtomicLong lastTick = new AtomicLong(0L);
    private Subscription subscription;

    void resume() {
        System.out.println("resumed");
        subscription = Observable.interval(5, TimeUnit.SECONDS, Schedulers.io())
                                 .map(tick -> lastTick.getAndIncrement())
                                 .subscribe(tick -> System.out.println("tick = " + tick));
    }

    void stop() {
        if (subscription != null && !subscription.isUnsubscribed()) {
            System.out.println("stopped");
            subscription.unsubscribe();
        }
    }
}

Answer 2

前段时间，我也在寻找RX＆＃34;计时器＆＃34;解决方案，但没有满足我的期望。所以你可以找到我自己的解决方案：

AtomicLong elapsedTime = new AtomicLong();
AtomicBoolean resumed = new AtomicBoolean();
AtomicBoolean stopped = new AtomicBoolean();

public Flowable<Long> startTimer() { //Create and starts timper
    resumed.set(true);
    stopped.set(false);
    return Flowable.interval(1, TimeUnit.SECONDS)
            .takeWhile(tick -> !stopped.get())
            .filter(tick -> resumed.get())
            .map(tick -> elapsedTime.addAndGet(1000));
}

public void pauseTimer() {
    resumed.set(false);
}

public void resumeTimer() {
    resumed.set(true);
}

public void stopTimer() {
    stopped.set(true);
}

public void addToTimer(int seconds) {
    elapsedTime.addAndGet(seconds * 1000);
}

Answer 3

val switch = new java.util.concurrent.atomic.AtomicBoolean(true)
val tick = new java.util.concurrent.atomic.AtomicLong(0L)

val suspendableObservable = 
  Observable.
    interval(5 seconds).
    takeWhile(_ => switch.get()).
    repeat.
    map(_ => tick.incrementAndGet())

您可以将switch设置为false以暂停滴答，并true将其恢复。

Answer 4

我认为这是另一种方法检查源代码时，您会发现使用 OnSubscribeTimerPeriodically 类 interval（）。关键代码如下。

@Override
public void call(final Subscriber<? super Long> child) {
    final Worker worker = scheduler.createWorker();
    child.add(worker);
    worker.schedulePeriodically(new Action0() {
        long counter;
        @Override
        public void call() {
            try {
                child.onNext(counter++);
            } catch (Throwable e) {
                try {
                    worker.unsubscribe();
                } finally {
                    Exceptions.throwOrReport(e, child);
                }
            }
        }

    }, initialDelay, period, unit);
}

所以，你会看到，如果你想要循环，那么在 onNext（）中抛出一个新的异常呢。下面的示例代码。

Observable.interval(1000, TimeUnit.MILLISECONDS)
            .subscribe(new Action1<Long>() {
                @Override
                public void call(Long aLong) {
                    Log.i("abc", "onNext");
                    if (aLong == 5) throw new NullPointerException();
                }
            }, new Action1<Throwable>() {
                @Override
                public void call(Throwable throwable) {
                    Log.i("abc", "onError");
                }
            }, new Action0() {
                @Override
                public void call() {
                    Log.i("abc", "onCompleted");
                }
            });

然后你会看到：

08-08 11:10:46.008 28146-28181/net.bingyan.test I/abc: onNext
08-08 11:10:47.008 28146-28181/net.bingyan.test I/abc: onNext
08-08 11:10:48.008 28146-28181/net.bingyan.test I/abc: onNext
08-08 11:10:49.008 28146-28181/net.bingyan.test I/abc: onNext
08-08 11:10:50.008 28146-28181/net.bingyan.test I/abc: onNext
08-08 11:10:51.008 28146-28181/net.bingyan.test I/abc: onNext
08-08 11:10:51.018 28146-28181/net.bingyan.test I/abc: onError

Answer 5

很抱歉这是在RxJS而不是RxJava，但概念将是相同的。我从learn-rxjs.io改编了这个版本，此处它位于codepen。

您的想法是，您开始使用两个点击事件流startClick$和stopClick$。 stopClick$流上发生的每次点击都会映射到一个空的observable，点击startClick$每个点都会映射到interval$流。两个结果流将merge - d组合成一个可观察的可观察量。换句话说，每次点击时，merge都会发出两种类型之一的新观察结果。得到的observable将通过switchMap，它开始收听这个新的observable并停止收听它之前听到的任何内容。 Switchmap还将开始将此新observable中的值合并到其现有流中。

切换后，scan只能看到＆＃34;增量＆＃34; interval$发出的值，当＆＃34;停止＆＃34;时，它看不到任何值。已被点击。

在第一次点击发生之前，startWith将开始从$interval发出值，只是为了让事情顺利进行：

const start = 0;
const increment = 1;
const delay = 1000;
const stopButton = document.getElementById('stop');
const startButton = document.getElementById('start');
const startClick$ = Rx.Observable.fromEvent(startButton, 'click');
const stopClick$ = Rx.Observable.fromEvent(stopButton, 'click');
const interval$ = Rx.Observable.interval(delay).mapTo(increment);
const setCounter = newValue => document.getElementById("counter").innerHTML = newValue;
setCounter(start);

const timer$ = Rx.Observable

    // a "stop" click will emit an empty observable,
    // and a "start" click will emit the interval$ observable.  
    // These two streams are merged into one observable.
    .merge(stopClick$.mapTo(Rx.Observable.empty()), 
           startClick$.mapTo(interval$))

    // until the first click occurs, merge will emit nothing, so 
    // use the interval$ to start the counter in the meantime
    .startWith(interval$)

    // whenever a new observable starts, stop listening to the previous
    // one and start emitting values from the new one
    .switchMap(val => val)

    // add the increment emitted by the interval$ stream to the accumulator
    .scan((acc, curr) => curr + acc, start)

    // start the observable and send results to the DIV
    .subscribe((x) => setCounter(x));

这是HTML

<html>
<body>
  <div id="counter"></div>
  <button id="start">
    Start
  </button>
  <button id="stop">
    Stop
  </button>
</body>
</html>

Answer 6

您可以使用takeWhile并循环执行，直到条件成立为止

Observable.interval(1, TimeUnit.SECONDS)
        .takeWhile {
            Log.i(TAG, " time " + it)
            it != 30L
        }
        .subscribe(object : Observer<Long> {
            override fun onComplete() {
                Log.i(TAG, "onComplete " + format.format(System.currentTimeMillis()))
            }

            override fun onSubscribe(d: Disposable) {
                Log.i(TAG, "onSubscribe " + format.format(System.currentTimeMillis()))
            }

            override fun onNext(t: Long) {
                Log.i(TAG, "onNext " + format.format(System.currentTimeMillis()))
            }

            override fun onError(e: Throwable) {
                Log.i(TAG, "onError")
                e.printStackTrace()
            }

        });

Answer 7

@AndroidEx，这是一个很好的答案。我做了一些不同的事情：

private fun disposeTask() {
    if (disposeable != null && !disposeable.isDisposed)
      disposeable.dispose()
  }

 private fun runTask() {
    disposeable = Observable.interval(0, 30, TimeUnit.SECONDS)
.flatMap {
        apiCall.runTaskFromServer()
.map{

when(it){
is ResponseClass.Success ->{
keepRunningsaidTasks()
}
is ResponseClass.Failure ->{
disposeTask() //this will stop the task in instance of a network failure.
}
}

}

如何停止和恢复Observable.interval发出滴答声

7 个答案: