Question

我想为JSON序列化定义一个类类，如下所示：

case class Foo(id: String, variety: Variety)

sealed trait Variety
case object Bar extends Variety { override def toString = "1" }
case object Baz extends Variety { override def toString = "2" }

它应该像这样序列化和反序列化：

write[Foo](Foo("foo1", Bar))
"""{"id":"foo1", "variety":"1"}"""

read[Foo]("""{"id":"foo2", "variety":"2"}""")
Foo("foo2", Baz)

但它序列化为{"id":"foo2", "variety":{}}并且无法反序列化。是否可以使用像这样的案例对象？我正在使用lift-json_2.10 2.6-M2？

使用lift-json处理枚举是否有更好的方法？

Answer 1

而不是案例对象，将Variety设为案例类，Bar / Baz val s

Answer 2

我使用以下类为case对象创建自定义序列化程序。如果您担心json的格式，只需选择fieldKey和identifierOverride以适合您的情况。

object SpecialObjectSerializers {
  val FieldKeyDefault = "scalaObject"

  class SingleObjectSerializer[T <: AnyRef](obj: T, FieldKey : String = FieldKeyDefault,
                                             identifierOverride : Option[JValue] = None) extends Serializer[T] {
    val Identifier = identifierOverride.getOrElse(JString(obj.getClass.getName))
    override def serialize(implicit format: Formats): PartialFunction[Any, JValue] = {
      case `obj` =>
        JObject(List(JField(FieldKey, Identifier)))
    }
    override def deserialize(implicit format: Formats): PartialFunction[(TypeInfo, JValue), T] = {
      case (_, JObject(List(JField(FieldKey, Identifier)))) => obj
    }
  }

  class RichSerializer[T](serializer: Serializer[T]) {
    def +(that: Serializer[T]) = {
      new Serializer[T] {
        override def deserialize(implicit format: Formats) = serializer.deserialize.orElse(that.deserialize)
        override def serialize(implicit format: Formats) = serializer.serialize.orElse(that.serialize)
      }
    }
  }

  implicit def toRichSerializer[T](serializer: Serializer[T]) = new RichSerializer(serializer)

  def caseObjectsSerializer[T <: AnyRef](objs: T*): Serializer[T] = {
    val serializers: Seq[Serializer[T]] = objs.map(new SingleObjectSerializer(_))
    serializers.reduceLeft[Serializer[T]] {
      case (s1, s2) => s1 + s2
    }
  }
}

案例对象的lift-json序列化

2 个答案: