java - RxJava2 - 在平行轨道上发出 onError 将得到 UndeliverableException

Question

尽管其中一个 rails 是发出的onError()，但isCancelled()在另一个 rails 中仍然返回 false，这会导致UndeliverableException. 如何检查下游是否取消平行轨道？

Disposable disposable = Flowable.create(new FlowableOnSubscribe<Integer>() {
    @Override
    public void subscribe(FlowableEmitter<Integer> emitter) throws Exception {
        System.out.println("Flowable.create-emitter.isCancelled:" + emitter.isCancelled());
        for (int i = 1; i < 10; i++) {
            emitter.onNext(i);
        }
        emitter.onComplete();
    }

}, BackpressureStrategy.BUFFER).parallel(6).runOn(Schedulers.io())
        .flatMap(new Function<Integer, Publisher<String>>() {
            @Override
            public Publisher<String> apply(Integer t) throws Exception {
                // TODO Auto-generated method stub
                return Flowable.create(new FlowableOnSubscribe<String>() {

                    @Override
                    public void subscribe(FlowableEmitter<String> emitter) throws Exception {
                            System.out.println("flatMap-before onError-isCancelled:" + emitter.isCancelled());
                            try {
                                if (true) { // assume trigger the error
                                    throw new Exception("Test");
                                }
                                if (!emitter.isCancelled()) {
                                    emitter.onNext(String.valueOf((t + 1)));
                                    emitter.onComplete();
                                }
                            } catch (Exception ex) {
                                if (!emitter.isCancelled()) {
                                    emitter.onError(ex);
                                }
                            }
                            System.out.println("flatMap-after onError-isCancelled:" + emitter.isCancelled());
                    }
                }, BackpressureStrategy.BUFFER);
            }

        }).sequential().subscribeOn(scheduler).observeOn(Schedulers.single())
        .subscribeWith(new ResourceSubscriber<String>() {

            public void onComplete() {
                System.out.println("onComplete");

            }

            public void onError(Throwable arg0) {
                System.out.println("onError:" + arg0.toString());

            }

            public void onNext(String arg0) {
                System.out.println("onNext:" + arg0);
            }

        });

标签： javarx-javarx-java2

Answer 1

我找到了解决方案。我需要添加一个全局错误消费者来解决问题。 https://github.com/ReactiveX/RxJava/wiki/What's-different-in-2.0#error-handling