将值传递给创建处理 websocket Play 2 的 Actor

Question

我通过以下方式接受 websocket 消息：

  def socket = WebSocket.tryAcceptWithActor[ChatCommand, JsValue] { request =>
    Future.successful(request.session.get("login") match {
      case None => Left(Forbidden)
      case Some(_) => Right(ChatActor.props)
    })
  }

如何将 "login" 的值传递给 actor？

编辑：

object ChatActor{
  def props(out: ActorRef) = Props(new ChatActor(out))
 ....
}

class ChatActor(out: ActorRef) extends Actor {

  def receive = ...

}

当我更换道具时，我无法在控制器中提供。

Answer 1

如果聊天演员是这样的

object ChatActor {
  def props(out: ActorRef, login: String): Props = Props(new ChatActor(out, login))
}

class ChatActor(out: ActorRef, login: String) extends Actor{
  override def receive: Receive = ???
}

然后你应该编写代码

  def socket = WebSocket.tryAcceptWithActor[ChatCommand, JsValue] { request =>
    Future.successful(request.session.get("login") match {
      case None => Left(Forbidden)
      case Some(login) => Right(out => ChatActor.props(out, login))
    })
  }

您可能会觉得有趣http://doc.akka.io/docs/akka/snapshot/scala/actors.html#Props

将值传递给创建处理 websocket Play 2 的 Actor

Pass value to creating an Actor handling websocket Play 2

scala

playframework

playframework-2.0