scala spray.json如何获取Json对象
如果我尝试Http响应{"ReturnValue":""}，这个代码会出错。
由: spray.json.DeserializationException:预期列表为JsArray引起，但得到{"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: $robots")
return "hello"
}
}
}由: spray.json.DeserializationException:预期列表为JsArray引起，但得到{"ReturnValue":""} at spray.json.package$.deserializationError(package.scala:23) at spray.json.CollectionFormats$$anon$1.read(CollectionFormats.scala:29) at spray.json.CollectionFormats$$anon$1.read(CollectionFormats.scala:25) at spray.httpx.SprayJsonSupport$$anonfun$sprayJsonUnmarshaller$1.applyOrElse(SprayJsonSupport.scala:37) at spray.httpx.SprayJsonSupport$$anonfun$sprayJsonUnmarshaller$1.applyOrElse(SprayJsonSupport.scala:34) at scala.runtime.AbstractPartialFunction.apply(AbstractPartialFunction.scala:36) at spray.httpx.unmarshalling.Unmarshaller$$anon$1$$anonfun$unmarshal$1.apply(Unmarshaller.斯卡拉:29)在spray.httpx.unmarshalling.SimpleUnmarshaller.protect(SimpleUnmarshaller.scala:40) at spray.httpx.unmarshalling.Unmarshaller$$anon$1.unmarshal(Unmarshaller.scala:29) at spray.httpx.unmarshalling.SimpleUnmarshaller.apply(SimpleUnmarshaller.scala:29) at spray.httpx.unmarshalling.SimpleUnmarshaller.apply(SimpleUnmarshaller.scala:23) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon$3.apply(UnmarshallerLifting.scala:35) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon$3.apply(UnmarshallerLifting.斯卡拉:34)在spray.httpx.unmarshalling.UnmarshallerLifting$$anon$2.apply(UnmarshallerLifting.scala:30) at spray.httpx.unmarshalling.UnmarshallerLifting$$anon$2.apply(UnmarshallerLifting.scala:29) at spray.httpx.unmarshalling.package$PimpedHttpResponse.as(package.scala:51) at spray.httpx.ResponseTransformation$$anonfun$unmarshal$1.apply(ResponseTransformation.scala:33) . 13
有办法得到Json对象吗？
回答 1
Stack Overflow用户
发布于 2017-12-13 01:48:05
您可以提供和使用您自己的反封送处理实现，它将构造JsValue而不是ListRobot。JsValue将表示有效响应(机器人列表)或任意json响应(或者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)在将来(对管道的调用)返回JsValue之后，您可以尝试以受控的方式(例如在try块中)将其转换回ListRobot，并在失败时将其作为自定义的json响应处理。
https://stackoverflow.com/questions/47765257
复制相似问题
腾讯云开发者
