Question

我有四个Http Url，我需要一个一个地尝试。但是有一个问题，当其中一个url错误时，后面的url不被调用。伪代码是这样的： i = 0;

 Observable.fromArray("http://www.baidu.com/", "http://www.google.com/", "https://www.bing.com/")
                    .concatMap(new Function<String, ObservableSource<String>>() {
                        @Override
                        public ObservableSource<String> apply(String s) throws Exception {
                            return Observable.create(new ObservableOnSubscribe<String>() {
                                @Override
                                public void subscribe(ObservableEmitter<String> emitter) throws Exception {
                                    Log.i("settingsubscribe", i + "");
                                    if (i == 1) {
                                        emitter.onError(new Throwable("error"));

                                        //emitter.onComplete();
                                    } else {
                                        emitter.onNext(s.concat("_1"));
                                        emitter.onComplete();
                                    }
                                    i ++;
                                }
                            });
                        }
                    })

                    .subscribe(new Consumer<String>() {
                        @Override
                        public void accept(String s) throws Exception {
                            Log.i("settingfffff", s);
                        }
                    }, new Consumer<Throwable>() {
                        @Override
                        public void accept(Throwable throwable) throws Exception {
                            Log.i("settingfffff", throwable.getMessage());
                        }
                    }, new Action() {
                        @Override
                        public void run() throws Exception {
                            Log.i("settingfffff", "onComplete");
                        }
                    });

当我= 1时，发送错误并且事件停止，我希望继续，该怎么办？

Answer 1

emitter.onError(new Throwable("error"));

在上面的代码行onError中，您正在创建新的Throwable，而是可以记录错误

关于rxjava请求网址

1 个答案: