使用 json4s 反序列化嵌套的多态 json 字段答案

【问题标题】：Deserialize nested polymorphic json field with json4s使用 json4s 反序列化嵌套的多态 json 字段
【发布时间】：2015-10-29 11:35:39
【问题描述】：

我有一个常见问题，但仍然无法理解我正在阅读的内容。

在一个 scalatra 应用程序中，我收到以下 json：

{
_type: "hello",
timestamp: 123,
data: [
{table: "stuffJ",_id: 24},
{table: "preferences",_id: 34,word: "john"}
]}

字段“数据”中的元素数量未知。字段表将始终存在以区分类类型。我正在尝试将其解析为 RestAPIMessage 类。这是我目前所拥有的：

    implicit val jsonFormats = new DefaultFormats { outer =>
        override val typeHintFieldName = "table"
        override val typeHints = ShortTypeHints(List(classOf[Preferences], classOf[StuffJ]))
      }

    sealed trait DataJson 
    case class Preferences(table: String, _id: Long, word : String) extends DataJson
    case class StuffJ(table: String, _id: Long) extends DataJson
    case class RestAPIMessage(_type: String, timestamp: Long, data: List[DataJson])

    // if sent as Json, returns a json with two "table" fields
    val message = new RestAPIMessage("hello", 123, List(new StuffJ("StuffJ", 24), new Preferences("preferences", 34, "john"))) 

    // if received as Json, fails with a "no usable value for outer" 
    val djson = """{"_type":"hello","timestamp":123,"data":[{"table":"StuffJ","_id":24},{"table":"table":"preferences","_id":34,"word":"john"}]}"""

感谢您的帮助！

【问题讨论】：

标签： scala json-deserialization json4s

【解决方案1】：

好吧，我想我明白了。最后，我似乎无法“开箱即用”我需要的东西，不得不编写一个自定义序列化程序。我找不到一个简单的“多态”示例，所以我在下面提供了一个最小的 foobar 示例：

import org.json4s.{DefaultFormats, Formats}
import org.json4s._
import org.json4s.JsonDSL._
import org.json4s.native.Serialization.{read, write}
import org.json4s.native.Serialization


sealed trait Bar 

case class Bar1(name : String, value: Int) extends Bar
case class Bar2(name : String, stuff: Int) extends Bar
case class Foo(timestamp:Long, bar: List[Bar])
var testFoo : Foo = new Foo(123, List(Bar1("bar1",1), Bar2("bar2",2)))

object BarSerializer extends CustomSerializer[Bar](format => (
  {
    case x: JObject =>
      val name = (x \ "name").extract[String]
      name match {
        case "bar1" =>
          val value = (x \ "value").extract[Int]
          Bar1(name,value)
        case "bar2" =>
          val value = (x \ "stuff").extract[Int]
          Bar2(name,value)
        case x => throw new MappingException("Can't convert bar with name " + x + " to Bar")
      }
  },
  {
    case x: Bar =>
      //if you need only the deserializing part above, I think you could replace the below with write(x)
      x match {
        case Bar1(a,b) => 
          ("name" -> a) ~ ("value" -> b)
        case Bar2(a,b) =>
          ("name" -> a) ~ ("stuff" -> b)
      }
  }
))

implicit val jsonFormats: Formats = Serialization.formats(NoTypeHints) + BarSerializer

val a = write(testFoo)
//returns : String = {"timestamp":123,"bar":[{"name":"bar1","value":1},{"name":"bar2","value":2}]}
read[Foo](a)
//returns : Foo = Foo(123,List(Bar1(bar1,1), Bar2(bar2,2)))

【讨论】：