swift – 结合两个Observables

我仍然是一个反应新手,我正在寻求帮助.

func doA() -> Observable<Void>
func dob() -> Observable<Void>

enum Result {
    case Success
    case BFailed
}

func doIt() -> Observable<Result> {

    // start both doA and dob. 
    // If both complete then emit .Success and complete
    // If doA completes,but dob errors emit .BFailed and complete
    // If both error then error

}

以上是我认为我想要的…初始函数doA()和dob()包装网络调用,因此它们将发出一个信号然后完成(或错误而不发出任何Next事件.)如果doA()完成但是dob()错误,我希望doIt()发出.BFailed然后完成.

感觉我应该使用zip或combineLatest,但我不知道如果我这样做,如何知道哪个序列失败了.我也很确定catchError是解决方案的一部分,但我不确定它究竟放在哪里.

–

正如我在想的那样,我对顺序发生的呼叫感到满意.那甚至可能更好……

IE：

Start doA() 
    if it completes start dob() 
        if it completes emit .Success 
        else emit .BFailed.
    else forward the error.

谢谢你的帮助.

解决方法

我相信.flatMapLatest()正是你要找的,链接你的可观察请求.

doFirst()
.flatMapLatest({ [weak self] (firstResult) -> Observable<Result> in
  // Assuming this doesn't fail and returns result on main scheduler,// otherwise `catchError` and `observeOn(MainScheduler.instance)` can be used to correct this
  // ...
  // do something with result #1
  // ...
  return self?.doSecond()
}).subscribeNext { [weak self] (secondResult) -> Void in
  // ...
  // do something with result #2
  // ...
}.adddisposableto(disposeBag)

这里是RxSwift中的.flatMapLatest()doc.

Projects each element of an observable sequence into a new sequence of observable sequences and then
transforms an observable sequence of observable sequences into an observable sequence producing values only from the most recent observable sequence. It is a combination of map + switchLatest operator.