Question

    Flux.create(sink -> {
        sink.error(new IOException());
        sink.error(new IOException());
    }).subscribe(System.out::println, System.out::println, System.out::println);

我（错误地）认为在第一次错误之后，磁通将处理掉而不再听取任何信号，但上面的代码会产生异常：

java.io.IOException
Exception in thread "main" reactor.core.Exceptions$BubblingException: java.io.IOException
    at reactor.core.Exceptions.bubble(Exceptions.java:154)
    at reactor.core.publisher.Operators.onErrorDropped(Operators.java:256)
    at reactor.core.publisher.FluxCreate$SerializedSink.error(FluxCreate.java:172)
    at main.App.lambda$f1$0(App.java:30)
    at reactor.core.publisher.FluxCreate.subscribe(FluxCreate.java:92)
    at reactor.core.publisher.Flux.subscribe(Flux.java:6447)
    at reactor.core.publisher.Flux.subscribeWith(Flux.java:6614)
    at reactor.core.publisher.Flux.subscribe(Flux.java:6440)
    at reactor.core.publisher.Flux.subscribe(Flux.java:6404)
    at main.App.f1(App.java:31)
    at main.App.main(App.java:23)
Caused by: java.io.IOException
    ... 8 more

为什么我得到这个例外？

Answer 1

根据反应流，两次发信号onError()是非法的，当create运算符收到第二个异常时，它会丢弃它。有一个钩子可以用来捕获这样的情况，并用它做一些有意义的事情，但在你的情况下，默认的实现开始，这只是包装无关的异常并抛出它，或者＆＃34; bubble＆＃34;它，给来电者。

在反应堆3中抛出多个错误

1 个答案: