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阅读 akka-streams 的文档，我不太清楚消息的顺序以及是否可以执行它。让我用我为聊天服务器编写的一小段代码来设置我的问题的上下文。

def flowShape(user: User) = GraphDSL
  .create(Source.actorRef[ChatMessage](bufferSize = 5, OverflowStrategy.fail)) {
    implicit builder =>
      implicit chatSource =>

      import GraphDSL.Implicits._

      val messageFromOutside = builder.add(Flow[String].map {
        case msg: String => UserTextMessage(user, msg)
        case _ => InvalidMessage
      })

      val merge = builder.add(Merge[ChatMessage](2))
      // UPDATE --> this is where the change comes
      // val merge = builder.add(Concat[ChatMessage](2))

      // val channelActorSink = Sink.actorRefWithAck(channelActor, ActorInitMessage, AckMessage, UserLeft(user))
      val channelActorSink = Sink.actorRef(channelActor, UserLeft(user))

      val actorAsSource = builder.materializedValue.map { actor => UserJoined(user, actor) }

      actorAsSource ~> merge.in(0)
      messageFromOutside.out ~> merge.in(1)
      merge ~> channelActorSink

      FlowShape(messageFromOutside.in, chatSource.out)
}

为了让我自己的事情变得简单，我将这个流形状与一个非常简单的源和汇一起使用。像这样的东西——

val source = Source(List[String]("hi", "hello", "what are you upto", "this is nice"))
val sink = Sink.foreach[ChatMessage] {
  case tm: UserTextMessage => println(s"${tm.user.username} :: ${tm.content}")
  case ul: UserLeft => println(s"${ul.user.username} just left the channel")
  case uj: UserJoined => println(s"${uj.user.username} just joined the channel")
  case _ => println(s"do not know what I just received")
}

val mychatchannel = new Channel(420, myactorsystem)

source.via(mychatchannel.chatFlow(User("sushruta"))).runWith(sink)

现在，我的担心来了。在终端中打印的事件顺序根本不对。而且我不确定如何解决它。这是我得到的输出——

[INFO] [11/10/2017 17:42:20.431] [akka-streams-akka.actor.default-dispatcher-5] [akka://akka-streams/user/channel-actor-420] sushruta sent a message
[INFO] [11/10/2017 17:42:20.441] [akka-streams-akka.actor.default-dispatcher-5] [akka://akka-streams/user/channel-actor-420] received a user joined message
[INFO] [11/10/2017 17:42:20.443] [akka-streams-akka.actor.default-dispatcher-5] [akka://akka-streams/user/channel-actor-420] sushruta sent a message
[INFO] [11/10/2017 17:42:20.444] [akka-streams-akka.actor.default-dispatcher-5] [akka://akka-streams/user/channel-actor-420] sushruta sent a message

hi输出中缺少第一条消息。该消息似乎在打印hi之前已发送。UserJoin message

actorRefWithAck我尝试通过使用（我在上面的代码中注释掉）来修复它（并在消息传递方面增加一些安全性） 。它给出了类似的输出。

[INFO] [11/11/2017 06:33:03.731] [akka-streams-akka.actor.default-dispatcher-3] [akka://akka-streams/user/channel-actor-420] channel initialized and ready to take events
[INFO] [11/11/2017 06:33:03.735] [akka-streams-akka.actor.default-dispatcher-3] [akka://akka-streams/user/channel-actor-420] sushruta sent a message
[INFO] [11/11/2017 06:33:03.736] [akka-streams-akka.actor.default-dispatcher-3] [akka://akka-streams/user/channel-actor-420] received a user joined message
[INFO] [11/11/2017 06:33:03.737] [akka-streams-akka.actor.default-dispatcher-4] [akka://akka-streams/user/channel-actor-420] sushruta sent a message
[INFO] [11/11/2017 06:33:03.737] [akka-streams-akka.actor.default-dispatcher-3] [akka://akka-streams/user/channel-actor-420] sushruta sent a message
[INFO] [11/11/2017 06:33:03.738] [akka-streams-akka.actor.default-dispatcher-3] [akka://akka-streams/user/channel-actor-420] sushruta sent a message
[INFO] [11/11/2017 06:33:03.738] [akka-streams-akka.actor.default-dispatcher-3] [akka://akka-streams/user/channel-actor-420] received a UserLeft message

显然，似乎正在发生的事情是源在发送消息之前发送UserJoin消息。我怎样才能解决这个问题？从概念上讲，我认为我希望在UserJoin message源实现但在它实际发送第一条消息之前立即发送。那可能吗？

谢谢

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1 回答 1

1

把溪流想象成水管：有水的时候，它就会流动。合并运算符不关心来自哪一侧的元素。如果你想订购这些输入，你需要使用 Concat 来告诉 Akka。

于 2017-11-12T07:22:11.823 回答