有没有办法实现以下目标?
我有两个Observable,
Observable<List<String>> broadcastIdsObservable
。Observable<List<EventData>> eventDatasObservable
。有没有办法用broadcastIdsObsevable中的List过滤eventDatasObservable?我需要List<EventData>
eventData
满足条件eventData.getBroadcastId().equals(broadCastId)
,其中broadCastId
是broadcastIds的每一项。
我尝试使用flatMap,map和Observable.concat的不同方法......但我无法实现它。
提前致谢。
答案 0 :(得分:0)
您可以使用RxJavaJoins。你需要包括
compile 'io.reactivex:rxjava-joins:0.22.0'
然后创建Join:
@NonNull
@SuppressWarnings("unchecked")
private Func0<Observable<List<EventData>>> loadAndFilterEvents() {
return new Func0<Observable<List<EventData>>>() {
@Override
public Observable call() {
Observable<List<String>> broadcastIdsObservable = ...;
Observable<List<EventData>> eventDatasObservable = ...;
Plan0<List<EventData>> plan = JoinObservable.from(broadcastIdsObservable)
.and(eventDatasObservable)
.then(filterJoin);
return JoinObservable.when(plan).toObservable();
}
private Func2<List<String>, List<EventData>, List<EventData>> filterJoin = (broadcastIds, events) -> {
List<EventData> filtered = new ArrayList<>();
for (EventData event : events) {
if (broadcastIds.contains(event.getBroadcastId())) {
filtered.add(event);
}
}
return filtered;
};
};
}
订阅它:
Observable.defer(loadAndFilterEvents())
.subscribeOn(Schedulers.io())
.observeOn(AndroidSchedulers.mainThread())
.subscribe(filteredEvents -> Timber.d("Events: %s", filteredEvents),
Timber::e);