Question

I’m trying to serialize some relatively simple models into json. For example, I’d like to get the json representation of:

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

Do i need to write my own Format[User] with the appropriate reads and writes methods or is there some other way? I’ve looked at https://github.com/playframework/Play20/wiki/Scalajson but I’m still a bit lost.

Answer 1

Yes, writing your own Format instance is the recommended approach. Given the following class, for example:

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

The instance might look like this:

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)
  ))
}

And you’d use it like this:

scala> User(1L, "Some", "Person", Some("[e-mail]"))
res0: User = User(1,Some,Person,Some([e-mail]))

scala> Json.toJson(res0)
res1: play.api.libs.json.JsValue = {"id":1,"firstName":"Some","lastName":"Person","email":"[e-mail]"}

scala> res1.as[User]
res2: User = User(1,Some,Person,Some([e-mail]))

See the documentation for more information.