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"优雅"`?
答案 0 :(得分:34)
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();
}
}
}
答案 1 :(得分:13)
前段时间,我也在寻找RX"计时器"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);
}
答案 2 :(得分:9)
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。
答案 3 :(得分:2)
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
答案 4 :(得分:1)
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>
答案 5 :(得分:0)
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()
}
});
答案 6 :(得分:0)
@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.
}
}
}