我试图将一个复杂的Scala模型/对象转换为Map[String，Any] Map的“值”可以是Map[String，Any]或String，Boolean，Int等。我在实际的模型中有多个层次的嵌套。我找不到一个干净的方法将任何给定的具有复杂嵌套的模型转换为Map[String，Any]
我尝试了下面的方法将模型转换为Map，但下面的方法不能正确地将内部模型转换为Map。
例如

def toMapWithOption[TCaseClass <: Product](tCaseClass: TCaseClass): Map[String, Any] = {
    tCaseClass.getClass.getDeclaredFields
      .map(_.getName)
      .zip(tCaseClass.productIterator.toIndexedSeq)
      .toSeq
      .toMap
      .filter { attribute =>
        attribute._2 match {
          case option: Option[Any] => option.isDefined
          case _                   => true
        }
      }
      .map { attribute =>
        attribute._2 match {
          case option: Option[Any] => attribute._1 -> option.get
          case _                   => attribute
        }
      }
  }

val contact_details = new ContactDetails(Some("00236485766"), "919347439483", "[email protected]")
val address = new Address("12B", "sample street", "Some_State", Some(contact_details))
val person = new Person(1, "John", address)
print(toMapWithOption(person));

case class Person(id: Int, name: String, address: Address)
case class ContactDetails(house_phone: Option[String], mobile_phone: String, email: String)
case class Address(house_number: String, street: String, state: String, contact_details: Option[ContactDetails])

字符串
实际响应：Map(id -> 1, name -> John, address -> Address(12B,sample street,Some_State,Some(ContactDetails(Some(00236485766),919347439483, [[email protected]](https://stackoverflow.com/cdn-cgi/l/email-protection) ))))
预期响应：Map(id -> 1, name -> John, address -> Map(house_number -> 12B, street -> sample street, state -> Some_State, contact_details -> Map(house_phone -> 00236485766, mobile_phone -> 919347439483, email -> [[email protected]](https://stackoverflow.com/cdn-cgi/l/email-protection) )))
仅供参考：没有任何嵌套的模型可以很好地使用上述方法进行转换