Question

Scala 2.12在这里使用Akka。当我的一个演员收到特定类型的消息（比如sentence = "This snowy weather is so cold" # Split the sentence into a list of the words words = sentence.split(" ") # Get the index of the word you want to replace word_to_replace_index = words.index("cold") # Replace the target word with the new word based on the index words[word_to_replace_index] = "awesome" # Generate a new sentence new_sentence = ' '.join(words)）时，我希望它将该消息转发给少数其他演员，完全按原样：

我试过了：

Fizzbuzz

虽然编译并没有抛出任何异常，但它不起作用（3个演员都没有收到class Foo extends Actor { override def receive: Receive = { case Bar => println("Bar!") case Whitstle => println("Whistle!") case Fizzbuzz => val actor1 = context.actorSelection("/user/a1") val actor2 = context.actorSelection("/user/a2") val actor3 = context.actorSelection("/user/a3") actor1 ! _ actor2 ! _ actor3 ! _ } }消息）。有什么想法吗？

Answer 1

在接收块中，收集变量中的msg，然后将该msg转发给其他actor。请参考以下代码： -

class Foo extends Actor {
  override def receive: Receive = {
    case Bar =>
      println("Bar!")
    case Whitstle =>
      println("Whistle!")
    case msg : Fizzbuzz =>
      val actor1 = context.actorSelection("/user/a1")
      val actor2 = context.actorSelection("/user/a2")
      val actor3 = context.actorSelection("/user/a3")

      actor1 ! msg 
      actor2 ! msg 
      actor3 ! msg 
  }
}

这应该可以解决您的问题。如果有任何疑问仍然存在，请告诉我。

转发给其他Akka演员的消息

1 个答案: