我怎么能抛出错误在onNext()与RxJava

Question

.subscribe(
    new Action1<Response>() {
        @Override
        public void call(Response response) {
            if (response.isSuccess())
            //handle success
            else
            //throw an Throwable(reponse.getMessage())
        }
    },
    new Action1<Throwable>() {
        @Override
        public void call(Throwable throwable) {
            //handle Throwable throw from onNext();
        }
    }
);

我不想在onNext()中处理(!response.isSuccess())。我怎么把它扔给onError()和其他可扔物一起处理?

Answer 1

如果FailureException extends RuntimeException，则

.doOnNext(response -> {
  if(!response.isSuccess())
    throw new FailureException(response.getMessage());
})
.subscribe(
    item  -> { /* handle success */ },
    error -> { /* handle failure */ }
);

如果你尽可能早地抛出异常，这样做效果最好，因为这样你就可以很容易地进行重试，替代响应等。

Answer 2

您可以对响应或错误的响应进行flatMap

flatMap(new Func1<Response, Observable<Response>>() {
    @Override
    public Observable<Response> call(Response response) {
        if(response.isSuccess()){
            return Observable.just(response);
        } else {
            return Observable.error(new Throwable(response.getMessage()));
        }
    }
})

Answer 3

解决方案是在中间添加一个操作符。我的建议是使用map，因为它不会生成新的Observable对象(与flatMap相比):

.map(new Func1<Response, Response>() {
    @Override
    public Response call(Response response) {
        if (response.isSuccess()) {
            return response;
        } else {
            throw new Throwable(reponse.getMessage()));
        }
    }
 })