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Monix:InputStreamObservable不支持多个订阅者

monix reactive-programming scala

4

N A · 技术社区 · 8 年前

我试图将(String,Date)的一个可观察对象拆分为两个不同的可观察对象,并将它们压缩在一起,如下所示

import monix.execution.Scheduler.Implicits.global
val x = Observable.fromIterator((0 to 10).map(i => (s"a $i", s"b $i")).toIterator)

val y = Observable.toReactive(x)

val fileStream = Observable.fromReactivePublisher(y).mapAsync(5)(a => Task{println(a._1); a._1})
val dateStream = Observable.fromReactivePublisher(y).mapAsync(5)(a => Task{println(a._2); a._2})

fileStream.zip(dateStream)
  .map(println)
  .subscribe()

但我有以下例外

monix.reactive.exceptions.MultipleSubscribersException: InputStreamObservable does not support multiple subscribers
    at monix.reactive.exceptions.MultipleSubscribersException$.build(MultipleSubscribersException.scala:51)
    at monix.reactive.internal.builders.IteratorAsObservable.unsafeSubscribeFn(IteratorAsObservable.scala:42)
    at monix.reactive.Observable$$anon$6.subscribe(Observable.scala:155)
    at monix.reactive.internal.builders.ReactiveObservable.unsafeSubscribeFn(ReactiveObservable.scala:38)
    at monix.reactive.internal.operators.MapAsyncParallelObservable.unsafeSubscribeFn(MapAsyncParallelObservable.scala:60)
    at monix.reactive.internal.builders.Zip2Observable.unsafeSubscribeFn(Zip2Observable.scala:158)
    at monix.reactive.Observable$$anon$5.unsafeSubscribeFn(Observable.scala:139)
    at monix.reactive.Observable$class.subscribe(Observable.scala:71)
    at monix.reactive.Observable$$anon$5.subscribe(Observable.scala:136)
    at monix.reactive.Observable$class.subscribe(Observable.scala:90)
    at monix.reactive.Observable$$anon$5.subscribe(Observable.scala:136)
    at monix.reactive.Observable$class.subscribe(Observable.scala:120)
    at monix.reactive.Observable$$anon$5.subscribe(Observable.scala:136)
    at monix.reactive.Observable$class.subscribe(Observable.scala:112)
    at monix.reactive.Observable$$anon$5.subscribe(Observable.scala:136)

2 回复 | 直到 8 年前

1

2

Sergei Shubin 8 年前

是否强制转换为/从被动?

解决这个问题的一个方法是 val x = Observable.fromIterable((0 to 10).map(i => (s"a $i", s"b $i"))) ,但对于无限个流,它将超出内存错误。

另一种方法是使用 .multicast(Pipe.publish[]) 然后 obs.connect() 记录代码:

import monix.execution.Scheduler.Implicits.global
val x = Observable.fromIterator((0 to 10).map(i => (s"a $i", s"b $i")).iterator)

val y = Observable.toReactive(x)
val obsY = Observable.fromReactivePublisher(y)
val connectY = obsY.multicast(Pipe.publish[(String, String)])

val fileStream = connectY.mapAsync(5)(a => Task{println(a._1); a._1})
val dateStream = connectY.mapAsync(5)(a => Task{println(a._2); a._2})

fileStream.zip(dateStream)
  .map(println)
  .subscribe()

connectY.connect()

Thread.sleep(5000)

2

0

Klugscheißer 7 年前

除了sergei shubin的回答之外,还可以暂时改变 Observable 进入“热”可观察状态,可使用 publishSelector 无需手动处理 multicast . 这看起来像:

val x = Observable.fromIterator((0 to 10).map(i => (s"a $i", s"b $i")).toIterator)

val zipped = x.publishSelector { o =>
  val fileStream = o.mapParallelUnordered(5)(a => Task{println(a._1); a._1})
  val dateStream = o.mapParallelUnordered(5)(a => Task{println(a._2); a._2})

  fileStream.zip(dateStream)
}

zipped
  .map(println)
  .subscribe()