如何在 RXSwift 中重置 Observable 间隔运算符？

Question

我是 RXSwift 的新手，我定义了一个 Observable 间隔计时器序列，它每秒调用一个 webservice 方法。在该网络服务响应中，我收到必须替换为我当前时间的新重试时间值。如何用新的时间值重置这个序列？这是我的代码：

func mySequence() {

    /////////////////////////////////////// subscribe to Timer (time change)
    var time = try! self.timer.value()

    self.disposeTimer = timer.subscribe({  value in

        time = value.element!

        print("=============================\(String(describing: time))=======================================")

    })
    /////////////////////////////////////// subscribe to Timer (time change)


    let   observable = Observable<Int>.interval(TIME I NEED TO CHANGE After response , scheduler: MainScheduler.instance).map { _ in ()
        self.myWebserviceMethod()
    }

    disposable =  observable.subscribe(onNext: {num in

    }, onError: { err in

    }, onCompleted: {

    }, onDisposed: {

    })

}

一切都发生了，但间隔计时器时间仍然是旧值:(

Answer 1

只需删除（处置）旧订阅并使用新间隔进行新订阅

var timerDisposable:Disposable?
var retryTime:RxTimeInterval = 1

func stratRefresh() {
    timerDisposable?.dispose()
    timerDisposable = Observable<Int>
        .timer(0, period: retryTime, scheduler: MainScheduler.instance)
        .subscribe(onNext: { value in
            myWebserviceMethod()
        })
}

func myWebserviceMethod() {
    // In service response update your retryTime
    // Ex.
    APIClient.getRetryTime() { newTime in
        if retryTime != newTime {
            retryTime = newTime
            stratRefresh()
        }
    }
}

如何在 RXSwift 中重置 Observable 间隔运算符？

How to reset the Observable interval operator in RXSwift?

ios

swift

rx-swift