Question

我正在开发一个使用actor来处理请求的播放应用程序。应用程序外线将是这样的..

MyController extends Controller {

  system.actorOf(Props(new MyRequestProcessor))
  val actorRef = MyRequestProcessor

  def handle(parse.json) { request =>
    actorRef ! request.body
    Ok
  }

}

基本上我想将MyRequestProcessor actor放在Supervisor actor下。由于控制器是由play框架创建的，我不知道在Supervisor中创建RequestProcessor并获取ActorRef的首选方法。

希望代码是这样的......

 Supervisor extends Actor {

   override val supevisorStrategy = OneForOneStrategy….

 }

 MyController extends Controller {

  val supervisorRef = system.actorOf(Props[Supervisor]...

  val processorRef = //NEED A WAY TO INSTANTIATE MyRequestProcessor WITHIN 
  //Supervisor and get that ActorRef back Using Supervisor's context.actorOf[]

 def handle(parse.json) { request
   processorRef ! request.body
 }

}

Answer 1

请问它。

import akka.actor._
import akka.pattern.ask
import scala.concurrent.Await
import scala.concurrent.duration._
import akka.util.Timeout

object Supervisor {
   case object GetChild
}

class Supervisor(childProps: Props) extends Actor {
   import Supervisor._

   val child = context.actorOf(childProps, name = "child")

   def receive = {
      case GetChild => sender ! child
   }
}

class Child extends Actor {
   def receive = Actor.emptyBehavior
}

object MyController extends Controller {
   val system = ActorSystem("my-system")
   val supervisor = system.actorOf(Props(classOf[Supervisor], Props[Child]), name = "parent")
   implicit val timeout = Timeout(5 seconds)
   // Blocking is not ideal, but is not the point in this example anyway
   val child = Await.ready((supervisor ? GetChild).mapTo[ActorRef], timeout.duration)
}

Play中的主管演员

1 个答案: