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在這個方法中,我希望看到返回的實際響應(result.toJson.toString或StatusCodes.InternalServerError.toString)而不是空字符串。我怎樣才能做到這一點?如何使方法在scala中同步?
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}"
}
}
這裏是代碼片段調用該方法,併發送響應於網頁套接字
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);
UPDATE 1:更新使用Await.result(今後,5000個米利斯)代替「未來的onComplete { ...}「。這是更改的代碼片段。現在工作,但只是想知道我們將如何處理失敗。
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
感謝。忘記.....現在固定 – 2015-01-20 18:58:14
改變了代碼到 響應= Await.result(今後,5000毫秒時間) 響應 獲取此編譯錯誤 [錯誤]實測值:scala.concurrent.Future [model.Registration ] [error] required:scala.concurrent.Awaitable [String] [error] response = Await.result(future,5000 millis) – 2015-01-20 19:10:01
你可以發佈你的新代碼來驗證,但看着錯誤,我猜測您正試圖將結果設置爲響應,而結果將返回註冊。我不知道這種類型,所以這就是爲什麼我只是把評論....我的猜測是它應該是結果(...)toJson.toString – 2015-01-20 19:29:55