Question

我创建了akka系统。假设其中有一些演员。我如何用他们的路径打印akka系统中的所有演员？（用于调试目的）

Answer 1

This reply by Roland Kuhn表明这不是一个完全无关紧要的问题，但您可以使用Identify-ActorIdentity请求 - 响应协议非常接近（对于将在合理时间内回复消息的演员）所有演员都服从。

将一些未经测试的代码拼凑在一起以说明这个想法：

import akka.actor._

def receive = {
    case 'listActors =>
      context.actorSelection("/user/*") ! Identify()

    case path: ActorPath =>
      context.actorSelection(path / "*") ! Identify()

    case ActorIdentity(_, Some(ref)) =>
      log.info("Got actor " + ref.path.toString)
      self ! ref.path
}

Answer 2

ActorSystem具有私有方法printTree，您可以将其用于调试。

1）私人方法来电者（来自https://gist.github.com/jorgeortiz85/908035）：

class PrivateMethodCaller(x: AnyRef, methodName: String) {
  def apply(_args: Any*): Any = {
    val args = _args.map(_.asInstanceOf[AnyRef])

    def _parents: Stream[Class[_]] = Stream(x.getClass) #::: _parents.map(_.getSuperclass)

    val parents = _parents.takeWhile(_ != null).toList
    val methods = parents.flatMap(_.getDeclaredMethods)
    val method = methods.find(_.getName == methodName).getOrElse(throw new IllegalArgumentException("Method " + methodName + " not found"))
    method.setAccessible(true)
    method.invoke(x, args: _*)
  }
}

class PrivateMethodExposer(x: AnyRef) {
  def apply(method: scala.Symbol): PrivateMethodCaller = new PrivateMethodCaller(x, method.name)
}

2）用法

val res = new PrivateMethodExposer(system)('printTree)()
println(res)

将打印：

-> / LocalActorRefProvider$$anon$1 class akka.actor.LocalActorRefProvider$Guardian status=0 2 children
   ⌊-> system LocalActorRef class akka.actor.LocalActorRefProvider$SystemGuardian status=0 3 children
   |   ⌊-> deadLetterListener RepointableActorRef class akka.event.DeadLetterListener status=0 no children
   |   ⌊-> eventStreamUnsubscriber-1 RepointableActorRef class akka.event.EventStreamUnsubscriber status=0 no children
   |   ⌊-> log1-Logging$DefaultLogger RepointableActorRef class akka.event.Logging$DefaultLogger status=0 no children
   ⌊-> user LocalActorRef class akka.actor.LocalActorRefProvider$Guardian status=0 1 children
...

要注意，它可能导致OOM如果你有很多演员。

Answer 3

根据the documentation，您可以将ActorSelection与通配符*一起使用，以使演员发送识别消息。你可以让一个演员收集ActorRef。

如@ chris-martin所述，只有当前不忙的演员才会发送。一个非常简单的演员：

// make all the available actor to send an identifying message
public void freeActors()
{
  ActorSelection selection =
    getContext().actorSelection("/user/*");
  selection.tell(new Identify(identifyId), getSelf());
}

...
// collect responses
@Override
public void onReceive(Object message) {
  if (message instanceof ActorIdentity) {
    ActorIdentity identity = (ActorIdentity) message;
    // get the ref of the sender 
    ActorRef ref = identity.getRef();
    // the sender is up and available
   ...

编辑：我知道这是针对Java的，但它似乎对我很有帮助。

如何打印akka系统中的所有演员？

3 个答案: