根据实例类型写入 JSON

Question

我在 Play for Scala 中有以下 class 结构：

sealed trait Father
final case class Child1 (name: String, descrip: String) extends Father
final case class Child2 (age: Int, price: Int) extends Father

case class MyClass (someValue: Int, father: Father)

现在，我需要为 MyClass:

定义隐式 JSON 函数

  implicit val myClassWrite : Writes[MyClass] = (
     (JsPath \ "some").write[Int] and
     (JsPath \ "father").write[Father]
  ) (unlift(MyClass.unapply))

我得到的是：

No Json serializer found for type Father. Try to implement an implicit Writes or Format for this type.

问题是无法为 Father 编写 Writes JSON 函数，因为它是没有属性的特征。我只能为 Child1 和 Child2 编写 JSON 函数，但我仍然得到错误。我的意图是让 Play 根据实例类型插入 Child1 或 Child2 的 JSON。 Play 可以做到这一点吗？

Answer 1

implicit val c1 = Json.format[Child1]
implicit val c2 = Json.format[Child2]

implicit val fatherWrites = new Writes[Father] {
  override def writes(o: Father) = o match {
    case x: Child1 => c1.writes(x)
    case x: Child2 => c2.writes(x)
  }
}

// 测试

val father : Father = Child1("aa","aaa")
Json.toJson(father).toString() // {"name":"aa","descrip":"aaa"}

根据实例类型写入 JSON

Writing JSON depending on instance type

scala

playframework

playframework-2.0

playframework-2.5