Akka http 丢失发件人参考

Question

收到 Future[httpResponse] 后，我试图将消息发送到 sender，但我丢失了 sender 的引用。

这是我的接收方法的代码：

    def receive = {
        case Seq(method: HttpMethod, endpoint: String, payload: String) ⇒ {
          // I have the correct sender reference
          implicit val materializer: ActorMaterializer = ActorMaterializer(ActorMaterializerSettings(context.system)) // needed by singleRequest method below
          // I have the correct sender reference

          val response: Future[HttpResponse] = Http(context.system).singleRequest(HttpRequest(method = method, uri = endpoint, entity = payload))
          println("http request sent")
          // I have the correct sender reference
          response onSuccess {
            case HttpResponse(statusCode, _, entity, _) ⇒ {
              entity.dataBytes.runFold(ByteString.empty)(_ ++ _).foreach { body ⇒ 
                // NO Reference to sender
                sender ! HttpConsumerResponse(statusCode = statusCode, contentType = entity.contentType, body = body.utf8String)
              }
            }
            case _ => println("http request success 2")
          }

          response onFailure {
            case exception: Throwable ⇒ {
              println("http request failure")
              throw exception
            } // Adopting let-it-crash fashion by re-throwning the exception
          }
        }
        case _ => println("I am httpConsumerActor and I don't know")
      }

如果我这样更改代码：

def receive = {
        case Seq(method: HttpMethod, endpoint: String, payload: String) ⇒ {
          // I have the correct sender reference
          implicit val materializer: ActorMaterializer = ActorMaterializer(ActorMaterializerSettings(context.system)) // needed by singleRequest method below
          // I have the correct sender reference

          val response: Future[HttpResponse] = Http(context.system).singleRequest(HttpRequest(method = method, uri = endpoint, entity = payload))
          println("http request sent")
          // I have the correct sender reference
          val mySender = sender
          response onSuccess {
            case HttpResponse(statusCode, _, entity, _) ⇒ {
              entity.dataBytes.runFold(ByteString.empty)(_ ++ _).foreach { body ⇒ 
                // NO Reference to sender
                mySender ! HttpConsumerResponse(statusCode = statusCode, contentType = entity.contentType, body = body.utf8String)
              }
            }
            case _ => println("http request success 2")
          }

          response onFailure {
            case exception: Throwable ⇒ {
              println("http request failure")
              throw exception
            } // Adopting let-it-crash fashion by re-throwning the exception
          }
        }
        case _ => println("I am httpConsumerActor and I don't know")
      }

一切正常，但我必须像这条线一样发送演员的参考，我知道这不是最好的方法：

val mySender = sender

Answer 1

你的拳头方法不起作用的原因是你是"closing over mutable state"，即sender()方法是在执行onComplete时执行的，并且不再包含引用。这是 Akka 中一个相当常见的错误，我们都曾遇到过！ :)

正如您自己发现的那样，正确的解决方案是预存储引用。还有其他选项，例如 "becoming" 其他选项，但为了您的使用，我会说预存储是 "nice" 和 "simple" 之间权衡的正确方法。

参考这些资源：, Blog post

Akka http 丢失发件人参考

Akka http losing sender reference

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akka

akka-http