如何在接收中记录内部参与者状态？答案

【问题标题】：How to log internal actor state in receive?如何在接收中记录内部参与者状态？
【发布时间】：2014-09-26 15:44:06
【问题描述】：

对于可以相当简洁地表达的 Actor，不得不添加块 ({...}) 只是为了添加日志命令，这令人沮丧。我想在处理消息之前记录我的内部状态，然后在处理消息之后记录 - 这可能吗？

def receive = {
    // I want to log here instead and remove the non-critical logs from below
    //  e.g. log.debug(s"Received $message")
    //       log.debug(s"Internal state is $subscriptions")
    case RegisterChannel(name, owner) => {
      getChannel(name) match {
        case Some(deadChannel: DeadChannel) => {
          subscriptions += (RealChannel(name, Server(owner)) -> subscriptions(deadChannel))
          subscriptions -= deadChannel
          context.watch(owner)
          log.debug(s"Replaced $deadChannel with a real channel $channels")
        }
        case Some(realChannel: RealChannel) =>
          log.error(s"Refusing to use RegisterChannel($name, $owner) due to $realChannel")
        case None => {
          subscriptions += (RealChannel(name, Server(owner)) -> Vector())
          context.watch(owner)
          log.debug(s"Registered a new channel $channels")
        }
      }
    }

    case Terminated(dead) => {
      getRole(dead) match {
        case Some(client: Client) => // Remove subscriptions
          log.debug(s"Received Client Terminated($dead) $client")
          subscriptionsFor(client).foreach { subscription =>
            subscriptions += (subscription._1 -> subscription._2.filterNot(c => c == client))
          }
        case Some(server: Server) => { // Remove any channels
          log.debug(s"Received Server Terminated($dead) $server")
          channelsBy(server).foreach { realChannel =>
            subscriptions += (DeadChannel(realChannel.name) -> subscriptions(realChannel))
            subscriptions -= realChannel
          }
        }
        case None =>
          log.debug(s"Received Terminated($dead) but no channel is registered")
      }
    }
    // I want to log here as well, to see what effect the message had
    //  e.g. log.debug(s"Finished $message")
    //       log.debug(s"Internal state is now $subscriptions")
}

我不确定这是 Akka 特定问题还是 Scala 模式匹配特定问题，所以我标记了两者

编辑：尝试@aepurniet 的答案后，我不知道如何解决编译器错误。接收需要返回PartialFunction[Any,Unit]，但是当匹配不是=> {...}中唯一的项目时，它似乎返回Any=>AnyRef

// Compiler error because msg=>{...} is not proper type
def receive = msg => { 
    log.info(s"some log")

    msg match {
      case RegisterChannel(name, owner) => {
        getChannel(name) match {
    <snip>

【问题讨论】：

标签： scala logging akka

【解决方案1】：

received = { case ... } 实际上是received = msg => msg match { case ... } 的简写。你可以重写 receive = msg => { log.info(..); msg match { case ... } } 你可能需要额外指定类型。

【讨论】：

这看起来很接近。当我尝试它时，我遇到了您预测的类型问题，但我不知道如何解决。我已经用最后一个小问题编辑了帖子，如果您再看一看，将不胜感激

【解决方案2】：

您可以像这样使用akka.event.LoggingReceive：

def receive = LoggingReceive {
  case ...
}

然后您将akka.actor.debug.receive 设置为on，这将记录（到调试）收到的所有消息以及它们是否被处理。

参见 Akka 官方文档中的Tracing Actor Invocations 部分。

对于额外的状态记录，您可以执行类似于LoggingReceive的操作

def withStateLogging(handler: Receive): Receive = {
  case msg if handler.isDefinedAt(msg) ⇒
    log.debug("before actual receive")
    log.debug(s"received: $msg")
    log.debug(s"state: $state")
    handler(msg)
    log.debug("after actual receive")
    log.debug(s"finished: $msg")
    log.debug(s"state: $state")
}

def receive = withStateLogging {
  case ...
}

【讨论】：

【解决方案3】：

编译器报错是因为 Actor#receive 返回类型是 Receive 实际定义为

type Receive = PartialFunction[Any, Unit]

这是一个很好的例子，如何使用可堆叠特征解决您的问题：how to add logging function in sending and receiving action in akka

这有点棘手，会覆盖PartialFunction 的默认行为。

【讨论】：