使用akka流时的事件顺序

Question

阅读akka流的文档，我不太清楚诸如消息顺序之类的东西，以及我是否可以执行它。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) 
} 

让事情简单我自己，我用这个流形，一个非常简单的源和宿。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，这里来了我的关注。打印在终端中的事件顺序根本不好。我不知道如何解决它。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 

输出中缺少第一条消息hi。在打印UserJoin message之前似乎已经发送了hi消息。

我试着通过使用actorRefWithAck（我在上面的代码中注释过）修复它（并且还添加了一些消息传递的安全性）。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 

显然似乎是发生的是，源发送邮件发送UserJoin消息之前。我怎样才能解决这个问题？从概念上讲，我认为我希望UserJoin message在资源实现时，但在实际发送第一条消息之前立即发送。那可能吗？

谢谢

Answer 1

把溪流想成水管：当有水时，它会流动。合并运算符不关心来自哪个元素。如果你想订购这些输入，你需要通过使用Concat来告诉Akka。