如何在 Play 框架中使用 Scala 将 JsValue 转换为模型 Class？

Question

我正在尝试设置从天气 API 获得的 JSON 响应以适应我定义的模型 class 以便轻松使用它，但是我做不到。

这是 class ：

import play.api.libs.json._
import play.api.libs.functional.syntax._

case class Forecast(var main: String, var description: String, var temp: Int, var tempMin: Int, var tempMax: Int)

object Forecast {
  implicit val forecastJsonFormat: Reads[Forecast] = (
      (JsPath \ "weather" \"main").read[String] and
        (JsPath \ "weather" \"description").read[String] and
        (JsPath \ "main" \"temp").read[Int] and
        (JsPath \ "main" \"temp_min").read[Int] and
        (JsPath \ "main" \"temp_max").read[Int]
    ) (Forecast.apply _)
}

这是控制器中的代码：

def weather = Action.async {
futureResponse.map(response => {
  val jsonString = response.json.toString()
  val jsonObject = Json.parse(jsonString)

  // TODO: Create t [Forecast] Object which represents the response.json data to send it to the view  below

  Ok(views.html.weather(t))
})}

我得到的 response.json 示例：

{"coord":{"lon":37.62,"lat":55.75},"weather":[{"id":600,"main":"Snow","description":"light snow","icon":"13n"},{"id":701,"main":"Mist","description":"mist","icon":"50n"}],"base":"stations","main":{"temp":269.15,"pressure":1024,"humidity":92,"temp_min":268.15,"temp_max":270.15},"visibility":3100,"wind":{"speed":2,"deg":200},"clouds":{"all":90},"dt":1546266600,"sys":{"type":1,"id":9029,"message":0.0029,"country":"RU","sunrise":1546235954,"sunset":1546261585},"id":524901,"name":"Moscow","cod":200}

Answer 1

您必须将 main 更改为 Seq[String]，将 description 更改为 Seq[String] 以及 temp、tempMin、tempMax至 Double

我在这里使用了不同的方式来创建reads，但是如果格式与预期格式不同，这种方式会抛出异常。

case class Forecast(main: Seq[String], description: Seq[String], temp: Double, tempMin: Double, tempMax: Double)

object Forecast {
    val reads = new Reads[Forecast] {
        override def reads(json: JsValue): JsResult[Forecast] = {
            val main = (json \ "weather" \ "main").map(_.as[String]).toList
            val description = (json \ "weather" \ "description").map(_.as[String]).toList
            val temp = (json \ "main" \ "temp").as[Double]
            val tempMin = (json \ "main" \ "temp_min").as[Double]
            val tempMax = (json \ "main" \ "temp_max").as[Double]

            JsSuccess(Forecast(main, description, temp, tempMin, tempMax))
        }
    }
}

或者您可以使用与您使用的相同的方式，但以不同的方式解析列表：

val forecastJsonFormat: Reads[Forecast] = (
  (JsPath \ "weather").read[List[Map[String, JsValue]]].map(_.map(_("main").as[String])) and
    (JsPath \ "weather").read[List[Map[String, JsValue]]].map(_.map(_("description").as[String])) and
    (JsPath \ "main" \ "temp").read[Double] and
    (JsPath \ "main" \ "temp_min").read[Double] and
    (JsPath \ "main" \ "temp_max").read[Double]
  ) (Forecast.apply _)

Answer 2

我终于做到了，方法如下：

在模型中，我定义了我的案例 class 和一个伴随对象，它解析我从网络 API 到我的 class 参数的 JSON 响应

型号代码：

import play.api.libs.json._
import play.api.libs.functional.syntax._

case class Forecast(main: String, description: String, temp: Double, tempMin: Double, tempMax: Double)

object Forecast {
  implicit val forecastReads: Reads[Forecast] = (
      (JsPath \ "weather" \ "main").read[String] and
      (JsPath \ "weather" \ "description").read[String] and
      (JsPath \ "main" \ "temp").read[Double] and
      (JsPath \ "main" \ "temp_min").read[Double] and
      (JsPath \ "main" \ "temp_max").read[Double]
    ) (Forecast.apply _)
}

在控制器代码中我添加了一个模式匹配，这里是！

控制器代码：

def weather = Action.async {
    futureResponse.map(response => {
      val parseResult = Json.fromJson[Forecast](response.json)
      parseResult match {
        case JsSuccess(forecast, JsPath) => Ok(views.html.weather(forecast))
        case JsError(error) => InternalServerError("Something went wrong!!") // Note that I'm not sure this result exists in Play...
      }
    })
  }

如何在 Play 框架中使用 Scala 将 JsValue 转换为模型 Class？

How to convert JsValue to a Model Class with Scala in Play framework?

scala

playframework

playframework-2.6