ios - Observable.empty 在 zip 中返回什么？

Question

如果一个函数返回一个 Observable.empty()，那么在执行这个值的 Observable.zip 之后我会得到什么？问题是我总是想执行 Observable.zip 中的代码，并且由于 source2() 可能会失败，所以我执行了一个 catchError 然后返回一个空的可观察对象。但我不确定 zip 是否仍会使用这种方法调用 block 。

 func source1() -> Observable<String> {
  return Observable.just("test")
}

func source2() -> Observable<Int> {
  return anObservableThatCanFail()
  .catchError { error -> Observable<Int> 
       return Observable.empty()
  }
}

func myFunc() {
    Observable.zip(source1(), source2()) { string, integer 
   //this will be called despite source2() do a empty()? 
   //and if so, what integer contains?
}

Answer 1

来自 reactivex.io的文档:

It applies this function in strict sequence, so the first item emitted by the new Observable will be the result of the function applied to the first item emitted by Observable #1 and the first item emitted by Observable #2; the second item emitted by the new zip-Observable will be the result of the function applied to the second item emitted by Observable #1 and the second item emitted by Observable #2; and so forth. It will only emit as many items as the number of items emitted by the source Observable that emits the fewest items.

所以在这种情况下，因为 .empty()将发出 0 项，意思是 zip也会发出 0 项。

如果你真的需要执行压缩功能，你可以改变source2()的类型来自 Observable<Int>至 Observable<Int?>并且，而不是返回 .empty()来自 catchError阻止，返回 .just(nil) .

func source2() -> Observable<Int?> {
  return anObservableThatCanFail()
    .map { (result: Int) -> Int? in result }
    .catchError { error -> Observable<Int?> 
       return Observable.just(nil)
    }
}