Question

我正在研究RX快速问题，以模拟4个用户点击。要求使它们异步发生以响应RX中的其他事件。所以我不能使用计时器或间隔。

我正在考虑一个函数，该函数将从可观察的对象中“拉出”一个可发出最多4个值然后终止的函数。我的问题是：

什么运算符允许我从头到尾“拉动”或逐步遍历可观察的所有元素？

<ul role="list">
  <li role="listitem">
    <div><h3 aria-label="item 1">item 1</h3></div>
  </li>
  <li role="listitem">
    <div><h3 aria-label="item 2">item 2</h3></div>
  </li>
  <li role="listitem">
    <div><h3 aria-label="item 2">item 3</h3></div>
  </li>
  <li role="listitem">
    <div><h3 aria-label="item 4">itwem 4</h3></div>
  </li>
</ul>

该想法是尝试在不依赖外部变量的情况下实现“纯” RX实现。我正在考虑func recursive(duration: int) -> Observable<Int> { // logic that may terminate recursion based on network conditions //logic to terminate if number of taps exceeded If I take from the taps array observable, and it completes - terminate recursion }，但正在努力查看它与解决方案的递归性质有何关系

Answer 1

如果我了解您的需求，几年前我使用诺言做了类似的事情。也许可以帮到您。 https://gist.github.com/dtartaglia/2b19e59beaf480535596

下面，我更新了承诺代码以使用Single：

/**
Repeatedly evaluates a promise producer until a value satisfies the predicate.
`promiseWhile` produces a promise with the supplied `producer` and then waits
for it to resolve. If the resolved value satisfies the predicate then the
returned promise will fulfill. Otherwise, it will produce a new promise. The
method continues to do this until the predicate is satisfied or an error occurs.

- Returns: A promise that is guaranteed to fulfill with a value that satisfies
the predicate, or reject.
*/

func doWhile<T>(pred: @escaping (T) -> Bool, body: @escaping () -> Single<T>, fail: (() -> Single<Void>)? = nil) -> Single<T> {
    return Single.create { event in
        func loop() {
            _ = body().subscribe(onSuccess: { (t) -> Void in
                if !pred(t) {
                    event(SingleEvent.success(t))
                }
                else {
                    if let fail = fail {
                        _ = fail().subscribe(onSuccess: { loop() }, onError: { event(SingleEvent.error($0)) })
                    }
                    else {
                        loop()
                    }
                }
            }, onError: {
                event(SingleEvent.error($0))
            })
        }
        loop()
        return Disposables.create()
    }
}

我不希望您能够使用上面的内容，但希望您能从中汲取灵感。

iOS RxSwift如何从Observable.of（1,2,3）中“拉”？

1 个答案: