如何以类型安全的方式映射play-ws异步Web请求的JSON响应?
private def webRequest(): Map[String, Any] = {
case class QuantileResult(val_05: Double, val_25: Double, val_50: Double, val_75: Double, val_90: Double)
object QuantileResult {
implicit val quantileResultFormat = Json.format[QuantileResult]
}
implicit val quantileReads = Json.reads[QuantileResult]
val payload = Json.obj(
"json" -> Json.parse("""{"type":1,"x":[1,2,3,4,5,6,7,8,9,10],"probs":[0.05,0.25,0.75,0.95]}"""),
"windowWidth" -> JsNumber(11)
)
wsClient.url("http://public.opencpu.org/ocpu/library/stats/R/quantile/json")
.withHeaders("Content-Type" -> "application/json")
.post(payload)
.map { wsResponse =>
if (!(200 to 299).contains(wsResponse.status)) {
sys.error(s"Received unexpected status, open-cpu error ${wsResponse.status} : ${wsResponse.body}")
}
println(s"OK, received ${wsResponse.body}")
wsResponse.json.validate[QuantileResult]
match {
case JsSuccess(arrOut: QuantileResult, _) => QuantileResult //TODO how to perform mapping here to Map[String, Any]
case e: JsError => JsError.toFlatForm(e) // TODO empty array does not work here otherwise future[object] does not conform to expected return type of Map[String, Any]
}
}
}
什么有效
wsClient.url("url").withHeaders("Content-Type" -> "application/json").post(payload).map { wsResponse => result = wsResponse.body }
val json = Json.parse(result)
val mapper = new ObjectMapper() with ScalaObjectMapper
mapper.registerModule(DefaultScalaModule)
val returnValue = mapper.readValue[Map[String, Any]](result)
但这似乎非常笨重,并不是异步。
答案 0 :(得分:1)
解决方案是将方法签名更改为适当的未来Future[Seq[Map[String, Any]]]
并通过Await.result(someMethod), 20.seconds)
解析
映射执行类似
wsResponse.json.validate[Seq[OutlierResult]] match {
case JsSuccess(result, _) => result.map(outlierRes => Map("period" -> outlierRes.period, "amount" -> outlierRes.amount, "outlier" -> outlierRes.outlier))
case JsError(error) => throw new OutlierParseException(error.toString())
}
即使这可以改进而不抛出异常