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如何使用先前可观察链的结果?

reactive rx-swift swift

13thang08 · 技术社区 · 6 年前

在一些平面图链之后,我希望最终使用前面的所有结果,假设我有以下代码:

func getObservable1() -> Observable<API1Response> {

    ...
}

func getObservable2(param: API1Response) -> Observable<API2Response> {

    ...
}

func getObservable3(param: API2Response) -> Observable<API3Response> {

    ...
}

getObservable1()
    .flatMap { api1Response in
        return getObservable2(api1Response) 
    }
    .flatMap { api2Response in
        return getObservable3(api2Response) 
    }
    .subscribe(onNext: { api3Response  in
        // I want to use both of api1Response and api2Response here
    })

我该怎么做?

我是否应该将api响应保存到out scope变量,然后使用后者?

类似于:

var api1Response: API1Response?
func getObservable1() -> Observable<API1Response> {
    ...

    self.api1Response = api1Response
    ...
}
...
getObservable1()
    ...
    .subscribe(onNext: { api3Response  in
        // use ap1Response here
        print(self.api1Response)
    })

谢谢

-----更新-----

基于@Misternewb answer,我发布了可运行的代码以供测试。

func getObservable1() -> Observable<Int> {
    return Observable.of(1)
}

func getObservable2(num: Int) -> Observable<Int> {
    return Observable.of(num + 1)
}

func getObservable3(num: Int) -> Observable<Int> {
    return Observable.of(num + 10)
}


getObservable1()
    .flatMap { ret1 in
        return getObservable2(num: ret1).map { (ret1, $0) }
    }
    .flatMap { ret1, ret2 in
        return getObservable3(num: ret2).map { (ret1, ret2, $0) }
    }
    .subscribe(onNext: { ret1, ret2, ret3 in
        print(ret1)
        print(ret2)
        print(ret3)
    })

输出

1
2
12

2 回复 | 直到 6 年前

Misternewb 6 年前

您可以尝试以下操作:

func getObservable1() -> Observable<API1Response> {
  return .empty()
}

func getObservable2(_ param: API1Response) -> Observable<API2Response> {
  return .empty()
}

func getObservable3(_ param: API2Response) -> Observable<API3Response> {
  return .empty()
}

func request() {
  getObservable1()
    .flatMap { api1Response in
      return getObservable2(api1Response).map { (api1Response, $0) }
    }
    .flatMap { api1Response, api2Response in
      return getObservable3(api2Response).map { (api1Response, api2Response, $0) }
    }
    .subscribe(onNext: { api1Response, api2Response, api3Response  in

    })
}

Mikhail Maslo 6 年前

您需要按照先决条件的顺序做出响应,还是只需要成功完成所有响应?

如果需要等待完成请求#1和请求#2,则可以使用

Observable.combineLatest(requestObservable1, requestObservable2)
  .flatMap { (requestResult1, requestResult2) -> Observable<You-Data-Type> in
     // Use result from request #1 & #2
     return requestObserver3
  }.subscribe(onNext: { (resultRequest3) in
  })
  ...