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以下代碼示例(可以複製並運行)顯示MyParentActor
,它創建一個MyChildActor
。Akka:Akka重新啓動後的消息排序
MyChildActor
爲其第一條消息引發異常,導致其重新啓動。
但是,我想要實現的是在「消息2」重新啓動MyChildActor
之前仍然處理「消息1」。
取而代之的是,消息1被添加到郵箱隊列的尾部,因此消息2被首先處理。
如何在重新啓動演員時獲得原始消息的排序,而無需創建自己的郵箱等?
object TestApp extends App {
var count = 0
val actorSystem = ActorSystem()
val parentActor = actorSystem.actorOf(Props(classOf[MyParentActor]))
parentActor ! "Message 1"
parentActor ! "Message 2"
class MyParentActor extends Actor with ActorLogging{
var childActor: ActorRef = null
@throws[Exception](classOf[Exception])
override def preStart(): Unit = {
childActor = context.actorOf(Props(classOf[MyChildActor]))
}
override def receive = {
case message: Any => {
childActor ! message
}
}
override def supervisorStrategy: SupervisorStrategy = {
OneForOneStrategy() {
case _: CustomException => Restart
case _: Exception => Restart
}
}
}
class MyChildActor extends Actor with ActorLogging{
override def preRestart(reason: Throwable, message: Option[Any]): Unit = {
message match {
case Some(e) => self ! e
}
}
override def receive = {
case message: String => {
if (count == 0) {
count += 1
throw new CustomException("Exception occurred")
}
log.info("Received message {}", message)
}
}
}
class CustomException(message: String) extends RuntimeException(message)
}