RxJava Observable 最小执行时间 [英] RxJava Observable minimum execution time

问题描述

我有一个 Observable(从网络获取数据).问题是 observable 可能快也可能慢，这取决于网络状况.

I have an Observable (which obtains data from network). The problem is that observable can be fast or slow depending on network conditions.

我在 observable 执行时显示进度小部件，并在 observable 完成时隐藏它.当网络速度快时 - 进度闪烁(出现和消失).我想将 observable 的最小执行时间设置为 1 秒.我该怎么做?

I show progress widget, when observable is executing, and hide it when observable completes. When the network is fast - progress flikers (appears and disappears). I want to set minimum execution time of observable to 1 second. How can I do that?

Delay"运算符不是一种选择，因为即使网络速度较慢，它也会延迟.

"Delay" operator is not an option because it will delay even for slow network.

推荐答案

你可以使用 Observable.zip() 来解决这个问题.给定

You can use Observable.zip() for that. Given

Observable<Response> network = ...

可以的

Observable<Integer> readyNotification = Observable.just(42).delay(1, TimeUnit.SECONDS);
Observable delayedNetwork = network.zipWith(readyNotification, 
                                                (response, notUsed) -> response);

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