Question

我已经创建了一条路线：

post {
        path("validate" / Segment / Segment / Segment) { (userId, listId, prefix) =>
            parameters('filter.?) { filter =>
                def result = (
                    phoneValidationActor ? ValidateUserList(userId.toInt, listId.toInt, prefix.toUpperCase, filter)
                )
                complete(result)
            }
        }
    }

还有演员

class PhoneNumberActor extends Actor with ActorLogging {

    import PhoneNumberActor._

    def receive: Receive = {
        case ValidateUserList(userId, listId, prefix, filter) =>
            sender() ! validateUserList(userId, listId, prefix, filter)
    }
}

还有演员的接收功能

def validateUserList(user_id: Int, list_id: Int, prefix: String, filter: Option[String]): Future[Seq[PhoneNumber]] = {
    val prefixTrim = prefix.trim
    val listContact = new ListContactRepository
    listContact.getAllContacts(user_id, list_id).map { lines =>
        lines.map { line =>
            validateNumber(line.phone, prefixTrim)
        }
    }
}

在路由中，结果被解析为Future[Any]而不是Future[Seq[PhoneNumber]] 需要帮助解决此问题

Answer 1

您需要致电询问电话mapTo。

(phoneValidationActor ? ValidateUserList(userId.toInt, listId.toInt, prefix.toUpperCase, filter))).mapTo[Seq[PhoneNumber]]

并在actor内部处理将来以避免ClassCastException

class PhoneNumberActor extends Actor with ActorLogging {

    import PhoneNumberActor._

    def receive: Receive = {
        case ValidateUserList(userId, listId, prefix, filter) =>
          val theSender = sender() //do not call this method in callback function onSuccess as it breaks actor principles and can send message to a wrong sender 
          validateUserList(userId, listId, prefix, filter).onSuccess { phoneNumbers => 
            theSender ! phoneNumbers
          }
    }
}

Answer 2

就像@Ivan Stanislavciuc所说的那样，您需要添加mapTo方法以将Future强制转换为期望的类型，在这种情况下为[Seq[PhoneNumber]]。对于您的路线，我会将其更新为类似这样的内容：

post {
    path("validate" / Segment / Segment / Segment) { (userId, listId, prefix) =>
        parameters('filter.?) { filter =>
            val result = 
                (phoneValidationActor ? ValidateUserList(userId.toInt, listId.toInt, prefix.toUpperCase, filter)).mapTo[Seq[PhoneNumber]]
            onSuccess(result) { maybeResult =>
              complete(maybeResult)
           }
        }
    }
}

如果您想处理成功和失败，则可以使用onComplete代替onSuccess，并具有以下内容：

                onComplete(result) {
                  case scala.util.Success(res) => complete(res)
                  case scala.util.Failure(ex) => complete(StatusCodes.BadRequest, ex)}

如何在akka-http中解决Future [Any]

2 个答案: