Question

我在Play Json for ID中有一个Json序列化器

  def idFormat[T] = {
    Format(new Reads[Id[T]] {
      def reads(jv: JsValue): JsResult[Id[T]] =
        JsSuccess(Id[T](jv.as[String]))
    }, new Writes[Id[T]] {
      def writes(id: Id[T]): JsString = JsString(id.underlying.toString)
    })
  }


  implicit def idFormatter[A]: Format[Id[A]] = idFormat[A]

现在，我想支持序列化Option [Id [_]]，但我的所有尝试都是徒劳的。我会怎么写

   implicit def optionIdFormatter[Option[A]]: ....

Answer 1

import play.api.libs.json._

case class Id[T](t: T)

implicit def idFormat[T](implicit tFormat: Format[T]): Format[Id[T]] = Format(
  new Reads[Id[T]] {
    def reads(jsValue: JsValue): JsResult[Id[T]] = jsValue.validate[T].map(t => Id(t))
  },
  new Writes[Id[T]] {
    def writes(id: Id[T]): JsValue = Json.toJson(id.t)
  }
)

implicit def optionFormat[T](implicit tFormat: Format[T]): Format[Option[T]] = Format(
  new Reads[Option[T]] {
    def reads(jsValue: JsValue): JsResult[Option[T]] = jsValue.validate[T].map(t => Some(t))
  },
  new Writes[Option[T]] {
    def writes(o: Option[T]): JsValue = o.map(t => Json.toJson(t)).getOrElse(JsNull)
  }
)

val intIdOpt = Some(Id(5))

val intIdOptJson = Json.toJson(intIdOpt)

val intIdOptJsonString = intIdOptJson.toString()
// intIdOptJsonString: String = 5

val usingOptionIntId1 = Map("id" -> Some(Id(5)))

val usingOptionIntId1JsonString = Json.toJson(usingOptionIntId1).toString()
// usingOptionIntId1JsonString: String = {"id":5}

val usingOptionIntId2 = Map("id" -> Option.empty[Id[Int]])

val usingOptionIntId2JsonString = Json.toJson(usingOptionIntId2).toString()
// usingOptionIntId2JsonString: String = {"id":null}

Play Json类型的选项[_]

1 个答案: