将有点复杂的类结构Scala到Json对象

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将其移动到json时。toJsonattach//attach是创建的附件对象，我得到错误：

AttachmentPayload.scala:18:未找到unapply或unapplySeq函数 [错误]隐式val attachmentPayloadWrites= Json.writes[AttachmentPayload]

我不知道如何创建unapply方法。

您的AttachmentPayload不是case类。您必须使其成为案例类：

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

或手动在伴生对象中创建应用/取消应用方法：

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，case类方法更简单，因此我建议您使用它，而不是手动创建apply/unapply方法