如何从广播的Akka流中获取订阅者和发布者？

debugcn 发表于 Dev

里普拉

使用更复杂的图形时，在使发布者和订阅者脱离流程时遇到了问题。我的目标是提供发布者和订阅者的API，并在内部运行Akka流。这是我的第一次尝试，效果很好。

val subscriberSource = Source.subscriber[Boolean]
val someFunctionSink = Sink.foreach(Console.println)

val flow = subscriberSource.to(someFunctionSink)

//create Reactive Streams Subscriber
val subscriber: Subscriber[Boolean] = flow.run()

//prints true
Source.single(true).to(Sink(subscriber)).run()

但是，对于更复杂的广播图，我不确定如何获取Subscriber和Publisher对象？我需要局部图吗？

val subscriberSource = Source.subscriber[Boolean]
val someFunctionSink = Sink.foreach(Console.println)
val publisherSink = Sink.publisher[Boolean]

FlowGraph.closed() { implicit builder =>
  import FlowGraph.Implicits._

  val broadcast = builder.add(Broadcast[Boolean](2))

  subscriberSource ~> broadcast.in
  broadcast.out(0) ~> someFunctionSink
  broadcast.out(1) ~> publisherSink
}.run()

val subscriber: Subscriber[Boolean] = ???
val publisher: Publisher[Boolean] = ???

鲁道夫

当您调用RunnableGraph.run()流时，运行结果是该运行的“物化值”。

在您简单的例子物化的价值Source.subscriber[Boolean]就是Subscriber[Boolean]。在您的复杂示例中，您希望将图形的几个组成部分的物化值合并为一个元组的物化值(Subscriber[Boolean], Publisher[Boolean])。

为此，您可以将您感兴趣的组件的物化值传递给它们FlowGraph.closed()，然后指定一个函数来组合物化值：

import akka.stream.scaladsl._
import org.reactivestreams._

val subscriberSource = Source.subscriber[Boolean]
val someFunctionSink = Sink.foreach(Console.println)
val publisherSink = Sink.publisher[Boolean]

val graph =
  FlowGraph.closed(subscriberSource, publisherSink)(Keep.both) { implicit builder ⇒
    (in, out) ⇒
      import FlowGraph.Implicits._

      val broadcast = builder.add(Broadcast[Boolean](2))

      in ~> broadcast.in
      broadcast.out(0) ~> someFunctionSink
      broadcast.out(1) ~> out
  }
val (subscriber: Subscriber[Boolean], publisher: Publisher[Boolean]) = graph.run()

有关FlowGraph.closed重载的更多信息，请参见Scaladocs。

（Keep.both是功能的缩写(a, b) => (a, b)）

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