用JSON4S反序列化Scala中的case对象
我有一些案例类的定义如下:用JSON4S反序列化Scala中的case对象,json,scala,scalatra,json4s,Json,Scala,Scalatra,Json4s,我有一些案例类的定义如下: sealed trait Breed case object Beagle extends Breed case object Mastiff extends Breed case object Yorkie extends Breed case class Dog(name: String, breed: Breed) 我还使用Scalatra定义了一个端点: post("/dog") { val dog = parsedBody.extract[Dog]
sealed trait Breed
case object Beagle extends Breed
case object Mastiff extends Breed
case object Yorkie extends Breed
case class Dog(name: String, breed: Breed)
我还使用Scalatra定义了一个端点:
post("/dog") {
val dog = parsedBody.extract[Dog]
...
}
我想要这个JSON对象:
{
name: "Spike",
breed: "Mastiff"
}
要反序列化到
狗的相应实例
。我正在努力弄清楚如何为Breed
编写自定义反序列化程序,并将其注册到JSON4S。您需要编写如下序列化程序:
序列化程序:
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("$",""))
}))
现在,您必须将这个序列化程序添加到默认格式
import org.json4s.CustomSerializer
val serializers = List(BreedSerializer)
implicit lazy val serializerFormats: Formats = DefaultFormats ++ serializers
希望这能解决您的问题。您可以为
品种
创建一个自定义序列化程序
,如您所述:
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")
}
))
您可以将其用作:
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)