Question

我遇到了Akka主管演员的问题。当子actor在未来结果的onFailure方法中抛出异常时，主管不处理错误（我想在ConnectException的情况下重启子进程）。

我正在使用Akka 2.3.7。

这是主管演员：

class MobileUsersActor extends Actor with ActorLogging {

  import Model.Implicits._
  import Model.MobileNotifications

  override val supervisorStrategy =
    OneForOneStrategy(maxNrOfRetries = 3, withinTimeRange = 1 minute) {
      case _: java.net.ConnectException => {
        Logger.error("API connection error. Check your proxy configuration.")
        Restart
      }
    }

  def receive = {
    case Start => findMobileUsers
  }

  private def findMobileUsers = {
    val notis = MobileNotificationsRepository().find()
    notis.map(invokePushSender)
  }

  private def invokePushSender(notis: List[MobileNotifications]) = {
    notis.foreach { n =>
      val pushSender = context.actorOf(PushSenderActor.props)
      pushSender ! Send(n)
    }
  }

}

这是儿童演员：

class PushSenderActor extends Actor with ActorLogging {

  def receive = {
    case Send(noti) => {
      val response = sendPushNotification(noti) onFailure {
        case e: ConnectException => throw e
      }
    }
  }

  private def sendPushNotification(noti: MobileNotifications): Future[WSResponse] = {
    val message = "Push notification message example"
    Logger.info(s"Push Notification >> $message to users " + noti.users)
    PushClient.sendNotification(message, noti.users)
  }

}

我尝试使用akka.actor.Status.Failure（e）通知发件人，如建议here，但是没有用，该例外由主管未处理。

作为一种解决方法，我找到了这种方法来实现它：

class PushSenderActor extends Actor with ActorLogging {

  def receive = {
    case Send(noti) => {
      val response = sendPushNotification(noti) onFailure {
        case e: ConnectException => self ! APIConnectionError
      }
    }
    case APIConnectionError => throw new ConnectException
  }

  private def sendPushNotification(noti: MobileNotifications): Future[WSResponse] = {
    val message = "Push notification message example"
    Logger.info(s"Push Notification >> $message to users " + noti.users)
    PushClient.sendNotification(message, noti.users)
  }

}

这是一个Akka错误还是我做错了什么？

谢谢！

Answer 1

我认为问题在于Future中抛出的异常与Actor运行的异常（可能）不属于同一个线程（更有经验的人可以详细说明）。因此，问题是Future体内抛出的异常被“吞噬”而不会传播到Actor。由于这种情况，Actor不会失败，因此不需要应用监督策略。所以，我想到的第一个解决方案是在一些消息中将异常包装在Future中，发送给自己，然后从Actor上下文中抛出它。这一次，将捕获异常并应用监督策略。但请注意，除非您再次发送Send（noti）消息，否则您将看不到自Actor重新启动以来发生的异常。总而言之，代码将是这样的：

class PushSenderActor extends Actor with ActorLogging {

  case class SmthFailed(e: Exception)

  def receive = {
    case Send(noti) => {
      val response = sendPushNotification(noti) onFailure {
        case e: ConnectException => self ! SmthFailed(e) // send the exception to yourself
      }
    }

    case SmthFailed(e) =>
      throw e // this one will be caught by the supervisor
  }

  private def sendPushNotification(noti: MobileNotifications): Future[WSResponse] = {
    val message = "Push notification message example"
    Logger.info(s"Push Notification >> $message to users " + noti.users)
    PushClient.sendNotification(message, noti.users)
  }

}

希望它有所帮助。

当child actor在未来的onFailure中抛出异常时，Akka主管演员不会处理异常

1 个答案: