Deserialization of case object in Scala with JSON4S

Question

I have some case classes defined like follows:

sealed trait Breed
case object Beagle extends Breed
case object Mastiff extends Breed
case object Yorkie exten         


        

Answer 1

You need to write the serializer like below:

Serializer:

case object BreedSerializer extends CustomSerializer[Breed](format => (
    {
      case JString(breed) =>  breed match {
        case "Beagle" => Beagle
        case "Mastiff" => Mastiff
        case "Yorkie" => Yorkie
      }
      case JNull => null
    },
    {
      case breed:Breed => JString(breed.getClass.getSimpleName.replace("$",""))
    }))

Now, you will have to add this serialiser to the default formats.

import org.json4s.CustomSerializer
val serializers = List(BreedSerializer)
implicit lazy val serializerFormats: Formats = DefaultFormats ++ serializers

Hope this solves your problem.

Answer 2

You can create a CustomSerializer for Breed like you mentioned :

import org.json4s._
import org.json4s.JsonDSL._
import org.json4s.native.JsonMethods._
import org.json4s.native.Serialization
import org.json4s.native.Serialization._

object BreedSerializer extends CustomSerializer[Breed]( format => ( 
  {
    case JString("Beagle")  => Beagle
    case JString("Mastiff") => Mastiff
    case JString("Yorkie")  => Yorkie
  }, {
    case Beagle  => JString("Beagle") 
    case Mastiff => JString("Mastiff")
    case Yorkie  => JString("Yorkie")
  }  
))

Which you can use as :

val json1 = """{ "name": "Spike", "breed": "Yorkie" }"""
val json2 = """{ "name": "Pluto", "breed": "Mastiff" }"""

implicit val json4sFormats = Serialization.formats(NoTypeHints) + BreedSerializer

val dog1 = parse(json1).extract[Dog] // Dog(Spike,Yorkie)
val dog2 = parse(json2).extract[Dog] // Dog(Pluto,Mastiff)