json - 使用 play.api.libs.json 将对象序列化为 json

Question

我正在尝试将一些相对简单的模型序列化为 json。例如，我想获得以下 json 表示:

case class User(val id: Long, val firstName: String, val lastName: String, val email: Option[String]) {
    def this() = this(0, "","", Some(""))
}

我是否需要使用适当的读取和写入方法编写自己的 Format[User] 还是有其他方法？我看过 https://github.com/playframework/Play20/wiki/Scalajson但我还是有点失落。

Answer 1

是的，自己写 Format实例是 the recommended approach .给定以下类，例如:

case class User(
  id: Long, 
  firstName: String,
  lastName: String,
  email: Option[String]
) {
  def this() = this(0, "","", Some(""))
}

该实例可能如下所示:

import play.api.libs.json._

implicit object UserFormat extends Format[User] {
  def reads(json: JsValue) = User(
    (json \ "id").as[Long],
    (json \ "firstName").as[String],
    (json \ "lastName").as[String],
    (json \ "email").as[Option[String]]
  )

  def writes(user: User) = JsObject(Seq(
    "id" -> JsNumber(user.id),
    "firstName" -> JsString(user.firstName),
    "lastName" -> JsString(user.lastName),
    "email" -> Json.toJson(user.email)
  ))
}

你会像这样使用它:

scala> User(1L, "Some", "Person", Some("s.p@example.com"))
res0: User = User(1,Some,Person,Some(s.p@example.com))

scala> Json.toJson(res0)
res1: play.api.libs.json.JsValue = {"id":1,"firstName":"Some","lastName":"Person","email":"s.p@example.com"}

scala> res1.as[User]
res2: User = User(1,Some,Person,Some(s.p@example.com))

见 the documentation想要查询更多的信息。