我对RXJava有点陌生。如果调用on error()而不丢失错误(我仍然希望对观察者调用on error(),那么我将尝试发出另一个项。但是,当我实现中声明的每个错误处理方法时,
docs
不会调用正在吞食和出错的错误。有什么解决办法吗?
编辑:
我昨天就是这么做的-
@Override
public Observable<ArrayList<Address>> getAirports() {
return new Observable<ArrayList<AirportPOJO>>() {
@Override
protected void subscribeActual(Observer<? super ArrayList<AirportPOJO>> observer) {
try {
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 (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 {
ArrayList<LocalDbAirportEntity> localAirportsEntities = localDb.getAirports();
ArrayList<Address> airports = AirportsMappers.getLocalDbAirportsToAirportsMapper().apply(localAirportsEntities);
return Observable.just(airports).concatWith(Observable.
<ArrayList<Address>>error(new Callable<Throwable>() {
@Override
public Throwable call() throws Exception {
return throwable;
}
}));
}
});
}
今天我想我可能做错了尝试-
@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;
}
但我也有同样的结果。使用第二个可观测的数据调用了onNext(),之后不调用onError()。