Question

我试图设置从天气API获得的JSON响应以适合我定义的模型类，以便轻松使用它，但我做不到。

这是课程：

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]，将temp，tempMin，tempMax更改为{ {1}}

我在这里使用了另一种创建Double的方法，但是如果格式与期望的格式不同，这种方法将引发异常。

reads

或者您可以使用相同的方式，但是以不同的方式解析列表：

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))
        }
    }
}

Answer 2

我终于做到了，这就是方法：

在模型中，我定义了我的案例类和一个伴侣对象，该对象解析从Web API到我的类参数的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Model类？

2 个答案: