如何讓websocket在play和Concurrent.broadcast上工作

Question

以下是document推薦的工作。

import play.api.mvc._ 
import play.api.libs.iteratee._ 
import play.api.libs.concurrent.Execution.Implicits.defaultContext 

def socket = WebSocket.using[String] { request => 

    // Concurrent.broadcast returns (Enumerator, Concurrent.Channel) 
    val (out, channel) = Concurrent.broadcast[String] 

    // log the message to stdout and send response back to client 
    val in = Iteratee.foreach[String] { 
    msg => println(msg) 
     // the Enumerator returned by Concurrent.broadcast subscribes to the channel and will 
     // receive the pushed messages 
     channel push("I received your message: " + msg) 
    } 
    (in,out) 
} 

但是它不工作，如果我更改爲：

def socket = WebSocket.using[String] { request => 

    val (out, channel) = Concurrent.broadcast[String] 

    val in=Iteratee.ignore[String] 
    channel push("Hello World") 

    (in,out) 
} 

我會很感激，如果你能幫助我理解爲什麼它不與新辦法的工作。

感謝

詹姆斯

更新：

class ServiceHandler extends Actor { 

    import Tcp._ 
    val (enumerator, channel) = Concurrent.broadcast[String] 

    val system=ActorDict.system 

    def receive = { 

    case subscribeData() =>{ 

     sender ! enumerator 


    } 

    case Received(data) => { 


     val dst = data.decodeString("utf-8") 
     val va=dst.substring(dst.lastIndexOf(',') + 1).trim() 
     println(va) 
      channel.push(va) 



    } 
    case PeerClosed => context stop self 
    } 

} 




    def ws = WebSocket.using[String] { request => 

    val in=Iteratee.ignore[String] 

    val dataHandler = Akka.system.actorOf(Props[ServiceHandler]) 

    val out= Await.result((dataHandler ? subscribeData()), 5 seconds).asInstanceOf[Enumerator[String]] 


    (in,out) 
    } 

Answer 1

您將不會收到這個「Hello World」的客戶端，因爲你彈到信道建立連接之前，與客戶。

如果您不需要接收來自客戶的任何東西（這就是爲什麼你在做val in=Iteratee.ignore[String]）。你最好和服務器發送的事件去：

Ok.chunked(out &> EventSource()).as("text/event-stream")