如何在scala中使方法同步？

Question

在这个方法中,我想看到实际的响应(result.toJson.toString或StatusCodes.InternalServerError.toString)返回而不是空字符串.我怎样才能做到这一点？

def process(msgIn : WebSocketMessageIn, service : ActorRef) : String = {
  import model.Registration
  import model.RegistrationJsonProtocol._

  implicit val timeout = Timeout(10 seconds)

  msgIn.method.toUpperCase match {
    case "POST" =>
      log.debug(s"Handing POST message with body ${msgIn.body}")
      val registration = msgIn.body.convertTo[Registration]
      val future = (service ? PostRegistrationMessage(registration)).mapTo[Registration]
      var response = ""
      future onComplete {
        case Success(result) =>
          response = result.toJson.toString

        case Failure(e) =>
          log.error(s"Error: ${e.toString}")
          response = StatusCodes.InternalServerError.toString
      }
      response

    case "PUT" =>
      s"Handing PUT message ${msgIn.body}"
  }
}

以下是调用该方法并将响应发送到websocket的代码片段

case Message(ws, msg, service) =>
  log.debug("url {} received msg '{}'", ws.getResourceDescriptor, msg)
  val wsMessageIn = msg.parseJson.convertTo[WebSocketMessageIn]
  val response = process(wsMessageIn, service)
  ws.send(response);

更新1:更新为使用Await.result(未来,5000毫秒)而不是'future onComplete {...}'.这是更改的代码段.现在工作,但只是想知道我们将如何处理失败.

msgIn.method.toUpperCase match {
  case "POST" =>
    log.debug(s"Handing POST message with body ${msgIn.body}")
    val registration = msgIn.body.convertTo[ADSRegistration]
    val future = (service ? PostADSRegistrationMessage(registration)).mapTo[ADSRegistration]
    val response = Await.result(future, 5000 millis)
    response.toJson.toString

Answer 1

您可以使用Await.result哪个阻止.像这样的东西:

import scala.concurrent.duration._
val result = Await.result(future, atMost = 10.second)
val response = //result processing

虽然如此,你可以通过未来的背部和执行send中onComplete这将是更加反应