每分钟rxjava重复一次观察的最佳方法

Question

我有以下方法:

public class ParentalControlInteractor {
   public Single<Boolean> isPinSet() {
       return bamSdk.getPinManager().isPINSet();
   }
}

我想调用此函数运行一次,然后每分钟重复一次,直到无穷大但这看起来很笨拙:

    parentalControlInteractor.isPinSet()
            .subscribeOn(Schedulers.io())
            .repeat(10000)
            .timeout(1600,TimeUnit.MILLISECONDS)
            .doOnError(throwable -> {
                Timber.e(throwable,"Error getting if Pin is set");
                throwable.printStackTrace();
            })
            .subscribe(isPinSet -> {
                this.isPinSet = isPinSet;
                Timber.d("Pin is set = " + isPinSet.toString());
                });

有没有更好的方法呢？我正在使用RxJava2.此外,上述方法仅调用10000次.我想永远调用它,就像使用Handler.postDelayed()一样.

Answer 1

您可以使用interval()oberator，这里是代码

DisposableObserver<Boolean> disposable = 
Observable.interval(1, TimeUnit.MINUTES)
            .flatMap(aLong -> isPinSet().toObservable())
            .subscribeOn(Schedulers.io())
            .subscribeWith({isPinSet -> doSomething()}, {throwable -> handleError()}, {});

如果您想随时完成此操作，请致电 disposable.dispose()

Answer 2

尝试 .repeatWhen(objectFlowable -> Flowable.timer(10, TimeUnit.SECONDS).repeat())

Answer 3

尝试这个：

parentalControlInteractor.isPinSet()
        .subscribeOn(Schedulers.io())
        .repeatWhen(new Func1<Observable<? extends Void>, Observable<?>>() {
            @Override
            public Observable<?> call(Observable<? extends Void> observable) {
                return observable.delay(60, TimeUnit.SECONDS);
            }
        })
        .doOnError(throwable -> {
            Timber.e(throwable,"Error getting if Pin is set");
            throwable.printStackTrace();
        })
        .subscribe(isPinSet -> {
            this.isPinSet = isPinSet;
            Timber.d("Pin is set = " + isPinSet.toString());
        });

Answer 4

事实证明这是在做这项工作：

parentalControlInteractor.isPinSet()
            .subscribeOn(Schedulers.io())
            .delay(10000,TimeUnit.MILLISECONDS)
            .repeat()
            .doOnError(throwable -> {
                Timber.e(throwable,"Error getting if Pin is set");
                throwable.printStackTrace();
            })
            .subscribe(isPinSet -> {
                this.isPinSet = isPinSet;
                Timber.d("Pin is set = " + isPinSet.toString());
                });

Answer 5

每次重复请求的最佳方式，第一次发射的特定延迟

 return Observable.interval(FIRST_ITEM_DELAY, CYCLE_TIME, TimeUnit.SECONDS)
                       .flatMap(aLong -> repository.repeatedRequest());