Scala有点复杂的类结构到Json对象

Question

我有这个结构

case class Attachment( Type: String = "template", var payload: AttachmentPayload){

}


object Attachment {
  implicit val attachmentWrites = Json.writes[Attachment]
}


object AttachmentPayload {

  implicit val attachmentPayloadWrites = Json.writes[AttachmentPayload]

}


class AttachmentPayload(val templateType: String, val elements: Option[ListBuffer[Element]] = None){

}

case class Element(title: String, imageUrl:String, subTitle: String, defaultAction: DefaultAction, buttons: Seq[Button]) {

}

当我尝试使用Json.toJson（attach）将其移动到json时// attach是创建的Attachment对象我得到错误：

  

AttachmentPayload.scala：18：找不到unapply或unapplySeq函数   [error]隐式val attachmentPayloadWrites =   Json.writes [AttachmentPayload]

我迷失了如何创建unapply方法。

Answer 1

您的AttachmentPayload不是案例类。您要么将其作为案例类：

case class AttachmentPayload(templateType: String, elements: Option[ListBuffer[Element]] = None)

或手动在随播对象中创建apply / unapplyMethods：

object AttachmentPayload {
  def apply(templateType: String, elements: Option[ListBuffer[Element]] = None): AttachmentPayload = new AttachmentPayload(templateType, elements)

  def unapply(value: Any): Option[(String, Option[ListBuffer[Element]])] = value match {
    case a: AttachmentPayload => Some((a.templateType, a.elements))
    case _ => None
  }
}

当然，案例类方法更简单，所以我建议您使用它而不是手动创建apply / unapply方法。