Question

我的服务中有2个单独的数据集合。

Featured和Standard内容。

我有2个api调用来返回这些项目。它们可以单独使用，但是当我要获取两组数据，根据条件提供一些丰富的数据然后将它们返回给消费者时，我也有用例。

我希望我可以做这样的事情：


class ContentService: ContentServiceType {

    let featured = PublishSubject<[Content]>()
    let standard = PublishSubject<[Content]>()

    let content: Observable<(featured: [Content], standard: [Content])>

    private let client: Client<ContentAPI>
    private let disposeBag = DisposeBag()

    init(client: Client<ContentAPI>) {
        self.client = client

        content = Observable
            .combineLatest(featured, standard)
            .map { (featured, standard) -> (featured: [Content], standard: [Content]) in

        /*
            Do some enrichment and create then return new, updated versions
        */

                return (featured: updatedFeatured, standard: updatedStandard)
        }.share()
    }

    func fetchStandardContent(page: Int = 0, size: Int = 100) -> Single<Void> {

        let params = ["page": page, "size": size]
        let request: Single<Content> = client.request(.getStandardContent(params))

        return request.map { [unowned self] launchers in
            self.standard.onNext(content.props)
            return ()
        }
    }

    func fetchFeaturedContent(page: Int = 0, size: Int = 100) -> Single<Void> {

        let params = ["page": page, "size": size]
        let request: Single<Content> = client.request(.getFeaturedContent(params))

        return request.map { [unowned self] content in
            self.featured.onNext(content.props)
            return ()
        }
    }
 }

然后在我的应用程序的其他地方，我希望可以做类似的事情

        contentSvc.content
            .observeOn(MainScheduler.instance)
            .subscribeOn(ConcurrentDispatchQueueScheduler(qos: .background))
            .subscribe(onNext: { content in
               /* do something w/ content */
            }).disposed(by: disposeBag)

然后，每当调用contentSvc.fetchFeaturedContent或contentSvc.fetchStandardContent时，上方的contentSvc.content订户都会获取新数据。

相反，content似乎没有发出任何值。

Answer 1

我相信

combineLatest要求两个源都先发射。

我也许会考虑使用BehaviorSubject或BehaviorRelay而不是PublishSubject。

Answer 2

我使用BehaviorRelay而不是PublishSubject，因为当从多个流绑定到PublishSubject（在应用之间共享）时，如果其中任何一个流发送完整消息，则PublishSubject可能会终止。中继类永远不会产生错误或永远不会完成。

let featured = BehaviorRelay(value: [Content]())
let standard = BehaviorRelay(value: [Content]())

func getContent() -> Observable<(featured: [Content], standard: [Content])> {
     return Observable
       .combineLatest(
         featured.asObservable(),
         standard.asObservable(),
         resultSelector: { (featured, standard) -> (featured: [Content], standard: [Content]) in
           return (featured: featured, standard: standard)
       }
     )
  }

func addElemToFeatured() {
    featured.accept([Content(name: "abc")])
  }

  func addElemToStandard() {
    standard.accept([Content(name: "xyz")])
  }

从初始化方法的不同类中调用getContent（）方法。还可以从按钮操作等不同位置调用addElemToFeatured，addElemToStandard。

listener!.getContent()
      .subscribe(onNext: { (featured, standard) in
        print(featured)
        print(standard)
      }).disposed(by: disposeBag)

合并两个PublishSubject的对象，并与Observable.combineLatest一起发射

2 个答案: