如何使用RxJava在onNext()中抛出错误

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();
        }
    }
);

我不想处理(!response.isSuccess())在onNext().我怎么能把它扔到onError()另一个扔掉的处理？

Answer 1

如果FailureException extends RuntimeException,那么

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

如果您尽早抛出异常,这种方法效果最好,因为您可以轻松地进行重试,替代响应等.

Answer 2

您可以flatMap对响应或错误做出回应

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()));
        }
    }
})