是的,自己写Format
实例是推荐的方法 http://www.playframework.com/documentation/2.0.1/ScalaJson。例如,给定以下类:
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)
))
}
你会这样使用它:
scala> User(1L, "Some", "Person", Some("[email protected] /cdn-cgi/l/email-protection"))
res0: User = User(1,Some,Person,Some([email protected] /cdn-cgi/l/email-protection))
scala> Json.toJson(res0)
res1: play.api.libs.json.JsValue = {"id":1,"firstName":"Some","lastName":"Person","email":"[email protected] /cdn-cgi/l/email-protection"}
scala> res1.as[User]
res2: User = User(1,Some,Person,Some([email protected] /cdn-cgi/l/email-protection))
See 文档 http://www.playframework.com/documentation/2.0.1/ScalaJsonGenerics了解更多信息。