Question

根据retryWhen() doc：

返回一个Observable，它发出与源相同的值除了onError之外的observable。一个onError通知从源头将导致Throwable项目的发射 Observable作为notificationHandler的参数提供功能。如果Observable调用onComplete或onError，则重试将在子订阅上调用onCompleted或onError。除此以外，此Observable将重新订阅源Observable。

当count1值达到3时，将调用subscriber.onCompleted（）;在我的选项中，在调用subscriber.onCompleted（）和＆＃34; child_onCompleted＆之后，将调用子订阅者的onComplete（）方法。＃34;将出现在输出中，但它不是;为什么会发生这种情况？

  count1 = 0;  

      Observable.create(new Observable.OnSubscribe<Integer>() {
            @Override
            public void call(Subscriber<? super Integer> subscriber) {
                System.out.println("call_onErr ");
                subscriber.onError(new Throwable("gg!"));
            }
        }).retryWhen(new Func1<Observable<? extends Throwable>, Observable<?>>() {
            @Override
            public Observable<?> call(Observable<? extends Throwable> observable) {

                System.out.println("fun_call+" + observable);
                return observable.flatMap(new Func1<Throwable, Observable<?>>() {
                    @Override
                    public Observable<?> call(Throwable throwable) {

                        if (count1 < 3) {
                            count1 = count1 + 1;
                            return Observable.create(new Observable.OnSubscribe<Integer>() {
                                @Override
                                public void call(Subscriber<? super Integer> subscriber) {
                                    System.out.println("fun_call_onNext " + "  count=" + count1);
                                    subscriber.onNext(1000);
                                }
                            });
                        } else
                            return Observable.create(new Observable.OnSubscribe<Integer>() {
                                @Override
                                public void call(Subscriber<? super Integer> subscriber) {
                                    System.out.println("fun_call_onCompleted " + "   " + count1);
                                    subscriber.onCompleted();//this is the subscriber!!!!!!!
                                }
                            });
                    }
                });


            }
        }).subscribe(new Subscriber<Integer>() {
            @Override
            public void onCompleted() {
                System.out.println("child_onCompleted ");
            }

            @Override
            public void onError(Throwable e) {
                System.out.println("child_err ");

            }

            @Override
            public void onNext(Integer integer) {
                System.out.println("child_onNext " + integer + " ");
            }
        });

输出是：

System.out﹕ fun_call+rx.Observable@3ecb7134
System.out﹕ call_onErr
System.out﹕ fun_call_onNext   count=1
System.out﹕ call_onErr
System.out﹕ fun_call_onNext   count=2
System.out﹕ call_onErr
System.out﹕ fun_call_onNext   count=3
System.out﹕ call_onErr
System.out﹕ fun_call_onCompleted    3

Answer 1

问题是你有flatMap，如果你发送一个空的Observable将无法完成。相反，您可以在其中使用特殊值并与takeWhile运算符通信以触发完成：

Observable.error(new Throwable("gg!"))
.retryWhen(o -> {
    AtomicInteger counter = new AtomicInteger();
    return o.flatMap(e -> {
        if (counter.getAndIncrement() < 3) {
            return Observable.just(0);
        }
        return Observable.just(1);
    })
    .takeWhile(v -> v == 0);
})
.subscribe(...);

在RxJava

1 个答案: