10

如何在 spray-json 中正确反序列化嵌套对象？

    import spray.json._

    case class Person(name: String)

    case class Color(n: String, r: Int, g: Int, b: Int, p: Person)

    object MyJsonProtocol extends DefaultJsonProtocol {

      implicit object ColorJsonFormat extends RootJsonFormat[Color] {
        def write(c: Color) = JsObject(
          "color-name" -> JsString(c.n),
          "Green" -> JsNumber(c.g),
          "Red" -> JsNumber(c.r),
          "Blue" -> JsNumber(c.b),
          "person-field" -> JsObject("p-name" -> JsString(c.p.name))
        )

        def read(value: JsValue) = {
          value.asJsObject.getFields("color-name", "Red", "Green", "Blue", "person-field") match {
            case Seq(JsString(name), JsNumber(red), JsNumber(green), JsNumber(blue), JsObject(person)) =>
              Color(name, red.toInt, green.toInt, blue.toInt, null) //gotta replace null with correct deserializer
            case _ => throw new DeserializationException("Color expected")
          }
        }
      }

    }

    import MyJsonProtocol._

    val jsValue = Color("CadetBlue", 95, 158, 160, Person("guest")).toJson

    jsValue.prettyPrint

    val color = jsValue.convertTo[Color] //person is missing of course

在旁注中，如何喷雾 json 帮助序列化到字段映射（嵌套对象的嵌套映射）？

4

3 回答 3

18

下面的示例演示 JSON -> Abstract Syntax Tree -> Scala Case Classes 并返回自定义字段名称和对可选案例类成员的支持。该示例源自https://github.com/spray/spray-json上 1.2.5 版的 spray-json 文档。

package rando

import spray.json._

case class Color(name: String, red: Int, green: Int, blue: Int)

case class Team(name: String, color: Option[Color])

object MyJsonProtocol extends DefaultJsonProtocol {
  implicit val colorFormat = jsonFormat(Color, "name", "r", "g", "b")
  implicit val teamFormat = jsonFormat(Team, "name", "jersey")
}
import MyJsonProtocol._

object GoSox extends App {
  val obj = Team("Red Sox", Some(Color("Red", 255, 0, 0)))
  val ast = obj.toJson
  println(obj)
  println(ast.prettyPrint)
  println(ast.convertTo[Team])
  println("""{ "name": "Red Sox", "jersey": null }""".asJson.convertTo[Team])
  println("""{ "name": "Red Sox" }""".asJson.convertTo[Team])
}

该示例在执行时输出以下内容：

Team(Red Sox,Some(Color(Red,255,0,0)))
{
  "name": "Red Sox",
  "jersey": {
    "name": "Red",
    "r": 255,
    "g": 0,
    "b": 0
  }
}
Team(Red Sox,Some(Color(Red,255,0,0)))
Team(Red Sox,None)
Team(Red Sox,None)
于 2014-02-12T05:24:49.100 回答
1

对于您剩下的问题 - 如何在包装类型中重用 JSON 转换：

"person-field" -> JsObject("p-name" -> JsString(c.p.name))

我会将其更改为：

"person-field" -> p.toJson

这样，您Person就让案例类 JSONify 自身，而不是在包装对象中引入另一种方式。干燥且更简单。

和：

case Seq(JsString(name), JsNumber(red), JsNumber(green), JsNumber(blue), JsObject(person)) => 
  Color(name, red.toInt, green.toInt, blue.toInt, null)

在.convertTo[Person]这里使用：

case Seq(JsString(name), JsNumber(red), JsNumber(green), JsNumber(blue), jsv) => 
  Color(name, red.toInt, green.toInt, blue.toInt, jsv.convertTo[Person])

如果有问题，请寻求更多帮助。我在自己的项目中有类似的结构，但没有尝试在这种情况下运行它们。

于 2015-07-02T08:03:14.947 回答
0 
import models.{Address, Patient}
import spray.json._

object Hospital {

  object MyJsonProtocol extends DefaultJsonProtocol {
    implicit val address = jsonFormat(Address, "country", "state", "zip")
    implicit val patient = jsonFormat(Patient, "name", "regNumber", "address")
  }

  import MyJsonProtocol._

  def main(args: Array[String]): Unit = {
    val p1 = Patient(name = "Amar", regNumber = 234, address = Address("IN", "KA", 49))
    println(p1.toJson.sortedPrint)
  }

}

输出

{
  "address": {
    "country": "IN",
    "state": "KA",
    "zip": 49
  },
  "name": "Amar",
  "regNumber": 234
}
于 2021-04-16T16:25:18.423 回答