ios - 链接不同类型的 RxSwift observable

Question

我需要从网络请求不同类型的模型，然后将它们组合成一个模型。 如何链接多个 observable 并返回另一个 observable？

我有这样的东西:

func fetchDevices() -> Observable<DataResponse<[DeviceModel]>>

func fetchRooms() -> Observable<DataResponse<[RoomModel]>>

func fetchSections() -> Observable<DataResponse<[SectionModel]>> 

我需要做类似的事情:

func fetchAll() -> Observable<(AllModels, Error)> {
    fetchSections()

    // Then if sections is ok I need to fetch rooms
    fetchRooms()

    // Then - fetch devices
    fetchDevices()

    // And if everything is ok create AllModels class and return it
    // Or return error if any request fails
    return AllModels(sections: sections, rooms: rooms, devices:devices)
  }

如何用RxSwift实现？我阅读了文档和示例，但了解如何链接具有相同类型的可观察对象

Answer 1

尝试 combineLatest 运算符。您可以组合多个可观察对象:

let data = Observable.combineLatest(fetchDevices, fetchRooms, fetchSections) 
    { devices, rooms, sections in
        return AllModels(sections: sections, rooms: rooms, devices:devices)
    }
    .distinctUntilChanged()
    .shareReplay(1)

然后，您订阅它:

data.subscribe(onNext: {models in 
    // do something with your AllModels object 
})
.disposed(by: bag)