Question

我对RxJava有点陌生。我正在尝试发出另一个项目，如果调用onError（）而不丢失错误（我仍然希望在观察者上调用onError（））。但是当我实现在docs中声明的每个错误处理方法时，该错误将被吞没，并且在发生错误时也不会被调用。有解决方案吗？

编辑：那就是我昨天试图做的-

@Override
public Observable<ArrayList<Address>> getAirports() {
    return new Observable<ArrayList<AirportPOJO>>() {
        @Override
        protected void subscribeActual(Observer<? super ArrayList<AirportPOJO>> observer) {
            try {
                // get airports from api list and map it
                ArrayList<AirportPOJO> airportsList = apiDb.getAirportsList(POJOHelper.toPOJO(AppCredentialManager.getCredentials()));
                observer.onNext(airportsList);
            } catch (Exception e) {
                e.printStackTrace();
                observer.onError(handleException(e));
            }
        }
    }.map(AirportsMappers.getAirportsPojoToDomainAirportsMapper()).doOnNext(new Consumer<ArrayList<Address>>() {
        @Override
        public void accept(ArrayList<Address> airportsList) throws Exception {
            // if airports loaded from api - save them to local db
            if (airportsList != null) {
                try {
                    localDb.saveAirportList(AirportsMappers.getAirportsToLocalDbAirportsMapper().apply(airportsList));
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }
    }).onErrorResumeNext(new Function<Throwable, ObservableSource<? extends ArrayList<Address>>>() {
        @Override
        public ObservableSource<? extends ArrayList<Address>> apply(final Throwable throwable) throws Exception {
            // load the local airports - 
            ArrayList<LocalDbAirportEntity> localAirportsEntities = localDb.getAirports();
            // map
            ArrayList<Address> airports = AirportsMappers.getLocalDbAirportsToAirportsMapper().apply(localAirportsEntities);
            // return the concat observable with the error
            return Observable.just(airports).concatWith(Observable.
                    <ArrayList<Address>>error(new Callable<Throwable>() {
                        @Override
                        public Throwable call() throws Exception {
                            return throwable;
                        }
                    }));
        }
    });
}

今天我坚强地做错了并尝试过-

@Override
public Observable<ArrayList<Address>> getAirports() {
    ArrayList<Observable<ArrayList<Address>>> observables = new ArrayList<>();
    observables.add(apiDb.getAirportsList(POJOHelper.toPOJO(AppCredentialManager.getCredentials())).map(AirportsMappers.getAirportsPojoToDomainAirportsMapper()));
    observables.add(localDb.getAirports().map(AirportsMappers.getLocalDbAirportsToAirportsMapper()));
    Observable<ArrayList<Address>> concatenatedObservable = Observable.concatDelayError(observables);
    return concatenatedObservable;
}

但是我得到了相同的结果。使用第二个可观察数据的onNext（）调用，之后不调用onError（）。

Answer 1

将期望值恢复为原始错误并恢复：

source.onErrorResumeNext(error ->
    Observable.just(item).concatWith(Observable.<ItemType>error(error))
);

RxJava错误处理可以消除错误

1 个答案: