Scala Akka http丢失发送方引用

在收到

未来[httpResponse]

后，我试图将消息发送给

发件人

，但我失去了

发件人的引用

以下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，但我必须发送演员的推荐信，就像这句话一样，我知道这不是最好的方式：

val mySender = sender

第一种方法不起作用的原因是您正在“关闭可变状态”，即执行onComplete时执行
sender（）
方法，并且不再包含引用。这在阿克卡是一个相当常见的错误，我们都有过

正确的解决方案，正如您自己已经发现的，是预先存储引用。还有其他选择，例如“成为”其他东西，但对于您的使用，我想说的是，就“好”和“简单”之间的权衡而言，pre-store是正确的方法

有关参考资料，请参阅以下参考资料：，
其他参考资料，Akka文档：