scala spray.json 如何获取 Json 对象

Question

如果我尝试 Http Get Response {"ReturnValue":""}，此代码出错。

Caused by: spray.json.DeserializationException: Expected List as JsArray, but got {"ReturnValue":""}

import spray.httpx.SprayJsonSupport._
import spray.json.DefaultJsonProtocol
import spray.http._
import spray.client.pipelining._
import scala.concurrent.duration._
import scala.concurrent.{ Await, Future }
import akka.actor.ActorSystem
import scala.concurrent.ExecutionContext.Implicits.global

class ApiHelper extends DefaultJsonProtocol {
  case class Robot(name: String, color: Option[String], amountOfArms: Int)

  implicit val RobotFormat = jsonFormat3(Robot)
  def CallAPI(httpMethod: String, subURL: String): String = {
    val apiLocation = "~~~"
    val timeout = 5.seconds

    implicit val system = ActorSystem("robotClient")
    return httpMethod match {
      case "GET" =>
        val pipeline: HttpRequest => Future[List[Robot]] = sendReceive ~> unmarshal[List[Robot]]
        val f: Future[List[Robot]] = pipeline(Get(s"$apiLocation"+subURL))
        val robots = Await.result(f, timeout)
        println(s"Got the list of robots: $robots")
        return "hello"
    }
  }
}

Caused by: spray.json.DeserializationException: Expected List as JsArray, but got {"ReturnValue":""} at

spray.json.package$.deserializationError(package.scala:23) at spray.json.CollectionFormats$$anon.read(CollectionFormats.scala:29) at spray.json.CollectionFormats$$anon.read(CollectionFormats.scala:25) at spray.httpx.SprayJsonSupport$$anonfun$sprayJsonUnmarshaller.applyOrElse(SprayJsonSupport.scala:37) at spray.httpx.SprayJsonSupport$$anonfun$sprayJsonUnmarshaller.applyOrElse(SprayJsonSupport.scala:34) at scala.runtime.AbstractPartialFunction.apply(AbstractPartialFunction.scala:36) at spray.httpx.unmarshalling.Unmarshaller$$anon$$anonfun$unmarshal.apply(Unmarshaller.scala:29) at spray.httpx.unmarshalling.SimpleUnmarshaller.protect(SimpleUnmarshaller.scala:40) at spray.httpx.unmarshalling.Unmarshaller$$anon.unmarshal(Unmarshaller.scala:29) at spray.httpx.unmarshalling.SimpleUnmarshaller.apply(SimpleUnmarshaller.scala:29) at spray.httpx.unmarshalling.SimpleUnmarshaller.apply(SimpleUnmarshaller.scala:23) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon.apply(UnmarshallerLifting.scala:35) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon.apply(UnmarshallerLifting.scala:34) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon.apply(UnmarshallerLifting.scala:30) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon.apply(UnmarshallerLifting.scala:29) at spray.httpx.unmarshalling.package$PimpedHttpResponse.as(package.scala:51) at spray.httpx.ResponseTransformation$$anonfun$unmarshal.apply(ResponseTransformation.scala:33) ... 13 more

有什么方法可以得到Json对象吗？

Answer 1

您可以提供并使用您自己的 unmarshal 实现，它将构造 JsValue 而不是 List[Robot]。 JsValue 将代表有效响应（机器人列表）或任意 json 响应（或可能更多自定义对象类型）。

  def unmarshal: HttpResponse ⇒ JsValue =
    response ⇒
      if (response.status.isSuccess)
        response.as[List[Robot]] match {
          case Right(value) ⇒ value.toJson
          case Left(error: MalformedContent) ⇒
            response.as[JsObject] match {
              case Right(value) ⇒ value.toJson
              case Left(error)  => throw new PipelineException(error.toString)
            }

      case Left(error) ⇒ throw new PipelineException(error.toString)
    }
  else throw new UnsuccessfulResponseException(response.status)

在未来（调用管道）returns JsValue 之后，您可以尝试以受控方式（例如在 Try 块内）将其再次转换回 List[Robot] 并在失败的情况下处理它作为自定义 json 回复。

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