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如何在RXSwift中重置Observable interval運算符?

[英]How to reset the Observable interval operator in RXSwift?

我是RXSwift的新手,我定義了一個Observable interval定時器序列,該序列每秒調用一次Webservice方法。 在該Web服務響應中,我收到新的重試時間值,該值必須替換為當前時間。 如何用新的時間值重置此序列? 這是我的代碼:

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: {

    })

}

一切都會發生,但是間隔計時器時間仍然是舊值:(

只需刪除(處置)舊訂閱並以新間隔創建新訂閱

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()
        }
    }
}

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