scala - Akka http 丢失发件人引用

Question

收到 Future[httpResponse] 后我正在尝试将消息发送到 sender但我失去了 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")
      }

如果我像这样更改代码:

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 Actor 的引用，我知道这不是最好的方法:

val mySender = sender

Answer 1

您的第一种方法不起作用的原因是您“关闭了可变状态”，即在执行 onComplete 时执行 sender() 方法，并且不再包含引用.这是 Akka 中一个相当常见的错误，我们都经历过! :)

正确的解决方案是预先存储引用，正如您自己已经发现的那样。还有其他选项，例如“成为”其他东西，但为了您的使用，我会说预存储是在“好”和“简单”之间权衡的正确方法。

有关引用，请参阅以下资源:SO question , Blog post