scala - 用列表 Play map 的scala json

Question

我有两个类，用户和地址。

case class User(
  id: Pk[Long] = NotAssigned,
  name: String = "",
  email: String = "",
  addresses: Seq[Address])

case class Address(
  id: Pk[Long] = NotAssigned,
  userId: Long,
  city: String)

我必须从我的 Controller 发送所有用户及其地址，例如 Map[User, List[Address]]。我可以使用 anorm (mysql) 提取它们，但随后我需要将它们作为 json 发送。能否请您帮助实现上述 Map[User, List[Address]] 的写入和读取，谢谢。

Answer 1

应该有帮助

import anorm._
import play.api.libs.json._
import play.api.libs.functional.syntax._


case class User(
  id: Pk[Long] = NotAssigned,
  name: String = "",
  email: String = "",
  addresses: Seq[Address])

case class Address(
  id: Pk[Long] = NotAssigned,
  userId: Long,
  city: String)

// Play does not provide Format[Pk[A]], so you need to define it
implicit def pkReads[A](implicit r: Reads[Option[A]]): Reads[Pk[A]] = r.map {  _.map(Id(_)).getOrElse(NotAssigned)  }
implicit def pkWrites[A](implicit w: Writes[Option[A]]): Writes[Pk[A]] =  Writes(id => w.writes(id.toOption))

implicit val addrFormat = Json.format[Address]
implicit val userFormat = Json.format[User]

现在您可以轻松地序列化用户:

val as = Seq(Address(Id(2), 1, "biim"))
val u = User(Id(1), "jto", "jto@foo.bar", as)

scala> Json.toJson(u)
res6: play.api.libs.json.JsValue = {"id":1,"name":"jto","email":"jto@foo.bar","addresses":[{"id":2,"userId":1,"city":"biim"}]}

正如 Julien 所说，您不能只序列化一个 Map[User, Seq[Address]]。它只是没有意义，因为 User 不能是 Json 对象中的键。