java-8 - RxJava 2.x 错误打印奇怪的堆栈跟踪

Question

我正在学习 RxJava(绝对是新手，抱歉，如果这个问题太基础了)，我在错误处理机制上遇到了困难(我已经阅读了文档，但这对我来说已经很先进了)。

这是我的代码，

public static void main(String[] args) {
    Observable<String> source = Observable.just("Alpha", "Beta", "Gamma", "Upma", "Idly");
    Observer<String> myObserver = new Observer<String>() {
      @Override
      public void onSubscribe(Disposable d) {
        // do nothing with Disposable, disregard for now
      }
  @Override
  public void onNext(String value) {
    System.out.println("RECEIVED: " + value);
    throw new RuntimeException("I am thrown");
  }

  @Override
  public void onError(Throwable e) {
    System.out.println("I got an error !");
    e.printStackTrace(new PrintStream(System.out));
  }

  @Override
  public void onComplete() {
    System.out.println("Done!");
  }
};
   source.subscribe(myObserver);
  }

这是我的堆栈跟踪

RECEIVED: Alpha
io.reactivex.exceptions.UndeliverableException: The exception could not be delivered to the consumer because it has already canceled/disposed the flow or the exception has nowhere to go to begin with. Further reading: https://github.com/ReactiveX/RxJava/wiki/What's-different-in-2.0#error-handling | java.lang.RuntimeException: I am thrown
    at io.reactivex.plugins.RxJavaPlugins.onError(RxJavaPlugins.java:367)
    at io.reactivex.Observable.subscribe(Observable.java:12275)
    at reactivex.ReactMain.main(ReactMain.java:36)
Caused by: java.lang.RuntimeException: I am thrown
    at reactivex.ReactMain$1.onNext(ReactMain.java:22)
    at reactivex.ReactMain$1.onNext(ReactMain.java:1)
    at io.reactivex.internal.operators.observable.ObservableFromArray$FromArrayDisposable.run(ObservableFromArray.java:108)
    at io.reactivex.internal.operators.observable.ObservableFromArray.subscribeActual(ObservableFromArray.java:37)
    at io.reactivex.Observable.subscribe(Observable.java:12268)
    ... 1 more
Exception in thread "main" io.reactivex.exceptions.UndeliverableException: The exception could not be delivered to the consumer because it has already canceled/disposed the flow or the exception has nowhere to go to begin with. Further reading: https://github.com/ReactiveX/RxJava/wiki/What's-different-in-2.0#error-handling | java.lang.RuntimeException: I am thrown
    at io.reactivex.plugins.RxJavaPlugins.onError(RxJavaPlugins.java:367)
    at io.reactivex.Observable.subscribe(Observable.java:12275)
    at reactivex.ReactMain.main(ReactMain.java:36)
Caused by: java.lang.RuntimeException: I am thrown
    at reactivex.ReactMain$1.onNext(ReactMain.java:22)
    at reactivex.ReactMain$1.onNext(ReactMain.java:1)
    at io.reactivex.internal.operators.observable.ObservableFromArray$FromArrayDisposable.run(ObservableFromArray.java:108)
    at io.reactivex.internal.operators.observable.ObservableFromArray.subscribeActual(ObservableFromArray.java:37)
    at io.reactivex.Observable.subscribe(Observable.java:12268)
    ... 1 more
Exception in thread "main" java.lang.NullPointerException: Actually not, but can't throw other exceptions due to RS
    at io.reactivex.Observable.subscribe(Observable.java:12277)
    at reactivex.ReactMain.main(ReactMain.java:36)
Caused by: java.lang.RuntimeException: I am thrown
    at reactivex.ReactMain$1.onNext(ReactMain.java:22)
    at reactivex.ReactMain$1.onNext(ReactMain.java:1)
    at io.reactivex.internal.operators.observable.ObservableFromArray$FromArrayDisposable.run(ObservableFromArray.java:108)
    at io.reactivex.internal.operators.observable.ObservableFromArray.subscribeActual(ObservableFromArray.java:37)
    at io.reactivex.Observable.subscribe(Observable.java:12268)
    ... 1 more

我有两个问题。

1) 我已经覆盖了 Observer 的 onError 方法。为什么我的 onError() 无法捕获异常？

2) 即使 onError 失败(我期待答案 1 中的原因)，为什么 UndeliverableException 只抛出两次？理想情况下，它必须被抛出 4 次，因为我有 4 个其他 Observable 字符串？

Answer 1

1.

来自 onError documentation :

Notifies the Observer that the Observable has experienced an error condition.

onError 将不会被调用，因为源可观察对象中没有错误。错误是在 observer 的 onNext 方法中引发的。如果你想测试 onError 你需要在流中抛出错误，例如:

sourece
    .map( str -> throw new RuntimeException("I am thrown: " + str))
    .subscribe(myObserver);

以上代码将调用 onError 而不是 onNext。

2。 为什么 UndeliverableException 只抛出两次？

我认为 UndeliverableException 只抛出一次，整个错误消息只描述了那一次崩溃。一旦您的代码在带有“alpha”的 onNext 方法中因错误退出，之后就不会发生任何事情。

尝试仅使用一个元素来运行您的代码，如下所示:

Observable<String> source = Observable.just("Alpha");

然后看看您是否收到相同的错误消息。此外，您还可以检查是否发出了任何东西:

Observable<String> source = Observable.just("Alpha", "Beta", "Gamma", "Upma", "Idly")
     .doOnNext(/* put log here to see what is being emitted */);