Question

我的申请必须做两件事：

同时只接受一个网络请求
请求失败时重试

这就是我实施它的方式：

public class RequestsLocker {

    private volatile boolean isLocked;

    public <T> Observable.Transformer<T, T> applyLocker() {
        if(!isLocked()) {
            return observable -> observable
                    .doOnSubscribe(() -> {
                        lockChannel();
                    })
                    .doOnUnsubscribe(() -> {
                        freeChannel();
                    });
        } else {
            return observable -> Observable.error(new ChannelBusyException("Channel is busy now."));
        }
    }

    private void lockChannel() {
        isLocked = true;
    }

    private void freeChannel() {
        isLocked = false;
    }

    public boolean isLocked() {
        return isLocked;
    }

}

看起来不错。

现在我的retryWhen实施：

public static Observable<?> retryWhenAnyIoExceptionWithDelay(Observable<? extends Throwable> observable) {
    return observable.flatMap(error -> {
        // For IOExceptions, we  retry
        if (error instanceof IOException) {
            return Observable.timer(2, TimeUnit.SECONDS);
        }

        // For anything else, don't retry
        return Observable.error(error);
    });
}

我有如何使用它：

public Observable<List<QueueCarItem>> finishService(int id, PaymentType paymentType, String notes) {
    return carsQueueApi.finishService(id, new FinishCarServiceRequest(paymentType.getName(), notes))
            .compose(requestsLocker.applyLocker(RequestsLocker.RequestChannel.CHANGE));
}

...

public void finishCarService(QueueCarItem carItem, PaymentType paymentType,
                             String notes, Subscriber<List<QueueCarItem>> subscriber) {
    queueApiMediator.finishService(carItem.getId(), paymentType, notes)
            .subscribeOn(ioScheduler)
            .observeOn(uiScheduler)
            .doOnError(this::handleError)
            .retryWhen(RxOperatorsHelpers::retryWhenAnyIoExceptionWithDelay)
            .subscribe(subscriber);
}

doOnUnsubscribe()调用任何错误然后锁定器的主要问题是对任何新请求打开，直到计时器到期并再次进行重新订阅。那就是问题所在。当计时器正在滴答作响时，用户可以再次提出请求。

我如何解决？

Answer 1

问题在于您将变压器应用于源可观察源，即retrywhen之前。如果出现错误，您将始终取消订阅，然后重新订阅源可观察对象导致doOnUnsubscribe被调用。

我建议你试试

public Observable<List<QueueCarItem>> finishService(int id, PaymentType paymentType, String notes) {
    return carsQueueApi.finishService(id, new FinishCarServiceRequest(paymentType.getName(), notes));            
}


public void finishCarService(QueueCarItem carItem, PaymentType paymentType,
                             String notes, Subscriber<List<QueueCarItem>> subscriber) {
    queueApiMediator.finishService(carItem.getId(), paymentType, notes)
            .subscribeOn(ioScheduler)
            .observeOn(uiScheduler)
            .doOnError(this::handleError)
            .retryWhen(RxOperatorsHelpers::retryWhenAnyIoExceptionWithDelay)
            .compose(requestsLocker.applyLocker(RequestsLocker.RequestChannel.CHANGE));
            .subscribe(subscriber);
}

PS：应用锁定器变换器看起来有点不同，即它不会在您链接的代码中使用参数。

Answer 2

使用retryWhen，为了避免取消订阅onError，您必须使用 onErrorResumeNext ，它不会取消订阅。

看看这个例子

/**
 * Here we can see how onErrorResumeNext works and emit an item in case that an error occur in the pipeline and an exception is propagated
 */
@Test
public void observableOnErrorResumeNext() {
    Subscription subscription = Observable.just(null)
                                          .map(Object::toString)
                                          .doOnError(failure -> System.out.println("Error:" + failure.getCause()))
                                          .retryWhen(errors -> errors.doOnNext(o -> count++)
                                                                     .flatMap(t -> count > 3 ? Observable.error(t) : Observable.just(null)),
                                                     Schedulers.newThread())
                                          .onErrorResumeNext(t -> {
                                              System.out.println("Error after all retries:" + t.getCause());
                                              return Observable.just("I save the world for extinction!");
                                          })
                                          .subscribe(s -> System.out.println(s));
    new TestSubscriber((Observer) subscription).awaitTerminalEvent(500, TimeUnit.MILLISECONDS);
}

关于并发性，如果在flatMap运算符中执行操作，则可以指定Max concurrent。

public final <R> Observable<R> flatMap(Func1<? super T, ? extends Observable<? extends R>> func, int maxConcurrent) {
    if (getClass() == ScalarSynchronousObservable.class) {
        return ((ScalarSynchronousObservable<T>)this).scalarFlatMap(func);
    }
    return merge(map(func), maxConcurrent);
}

您可以在此处查看更多示例https://github.com/politrons/reactive

Answer 3

我目前的解决方案不是在IoException上解锁RequestLocker，因为在这种情况下请求将在延迟后重复。

public <T> Observable.Transformer<T, T> applyLocker() {
    if(!isLocked()) {
        return observable -> observable.doOnSubscribe(() -> {
            lockChannel();
        }).doOnNext(obj -> {
            freeChannel();
        }).doOnError(throwable -> {
            if(throwable instanceof IOException) {
                return; // as any request will be repeated in case of IOException
            }
            freeChannel(channel);
        });
    } else {
        return observable -> Observable.error(new ChannelBusyException("Channel is busy now"));
    }
}

使用rxJava限制操作并重试操作符

3 个答案: