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publishSubject 的 onNext 方法正在连续调用（在不均匀的时间，大约在 1 毫秒内）并且要求每 1 秒发出这些项目，并且数据不应该丢失意味着应该发出每个项目。

    publishSubject.onNext("Data1");
    publishSubject.onNext("Data2");
    publishSubject.onNext("Data3");
    publishSubject.onNext("Data4");
    publishSubject.onNext("Data5");
    publishSubject.onNext("Data6");
    publishSubject.onNext("Data7");

等等...参见代码结构以供参考：

var publishSubject = PublishSubject.create<String>()
publishSubject.onNext(stateObject) // Executing at every milliseconds...


publishSubject
        /* Business Logic Required Here ?? */
        .subscribeOn(Schedulers.computation())
        .observeOn(AndroidSchedulers.mainThread())
        .subscribe {
            // Should execute at every 1 second
        }

请帮助，在此先感谢，

4

2 回答 2

1

这个类的功能扩展Observable正是您所需要的：

fun <T> Observable<T>.delayBetweenItems(timeout: Long, unit: TimeUnit): Observable<T> =
    Observable.zip(this, Observable.interval(timeout, unit), BiFunction<T, Long, T> { item, _ -> item })

您可以在项目中的某个实用程序类中声明它，然后像其他RxJava运算符一样应用它：

publishSubject
    .delayBetweenItems(1000, TimeUnit.MILLISECONDS)
    .subscribeOn(Schedulers.computation())
    .observeOn(AndroidSchedulers.mainThread())
    .subscribe {
        // Should execute at every 1 second
    }
于 2019-01-11T20:38:24.413 回答
1

将项目存储在 Deque 中怎么样？然后，使用启动挂起函数的协程每秒获取一次双端队列的第一个元素？

这是一些快速而肮脏的代码，只是为了确保它有效。您可以在kotlin 网站上在线运行此代码。请记住，我对 Kotlin 很陌生。

val deque: Deque<String> = ArrayDeque()
var refMillisAdd: Long = 0
var refMillisTake: Long = 0

fun main() {

    println(" Delay(ms) -> Action")
    println("---------------------")

    kotlinx.coroutines.runBlocking {

        launch {

            refMillisAdd = currentTimeMillis()
            refMillisTake = currentTimeMillis()

            for(i in 0..20){
               oncePer10ms(i.toString())
               refMillisAdd = currentTimeMillis()
            }

            for(i in 0..6){
                oncePerSecond()
                refMillisTake = currentTimeMillis()
            }
        }
    }
}

suspend fun oncePerSecond(){
    kotlinx.coroutines.delay(1_000L)
    println("  ${currentTimeMillis() - refMillisTake} -> TAKE ${deque.pop()}")
}

suspend fun oncePer10ms(item: String){
    kotlinx.coroutines.delay(10L)
    deque.add(item)
    println("  ${currentTimeMillis() - refMillisAdd} -> ADD $item")
}

上面的代码打印：

 Delay(ms) -> Action
---------------------
  17 -> ADD 0
  11 -> ADD 1
  10 -> ADD 2
  10 -> ADD 3
  10 -> ADD 4
  10 -> ADD 5
  10 -> ADD 6
  10 -> ADD 7
  10 -> ADD 8
  11 -> ADD 9
  10 -> ADD 10
  10 -> ADD 11
  10 -> ADD 12
  11 -> ADD 13
  10 -> ADD 14
  10 -> ADD 15
  11 -> ADD 16
  10 -> ADD 17
  10 -> ADD 18
  10 -> ADD 19
  11 -> ADD 20
  1223 -> TAKE 0
  1000 -> TAKE 1
  1000 -> TAKE 2
  1001 -> TAKE 3
  1000 -> TAKE 4
  1000 -> TAKE 5
  1000 -> TAKE 6
于 2019-01-11T16:48:50.813 回答