Question

收到Future[httpResponse]后，我尝试将邮件发送到sender但我忘记了sender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")
      }

如果我改变这样的代码：

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
          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，但我必须像这一行一样发送演员的引用，我知道这不是最好的方法：

val mySender = sender

Answer 1

你的第一种方法不起作用的原因是你正在关闭可变状态＆＃34;，即执行onComplete时执行sender()方法，并且不再包含引用。这是阿卡的一个相当常见的错误，我们都去过那里！：）

正确的解决方案，就像您已经发现的那样，是预先存储引用。还有其他选项，例如＆＃34;成为＆＃34;别的东西，但是为了你的使用我会说，预售是正确的方法，在＆＃34; nice＆＃34;之间进行权衡。和＆＃34;简单＆＃34;。

有关参考，请参阅以下资源：SO question，Blog post

Akka http丢失发件人参考

1 个答案: