使用play.api.libs.json将对象序列化为json [英] serializing objects to json with play.api.libs.json
问题描述
我正在尝试将一些相对简单的模型序列化为json.例如，我想获取以下内容的json表示形式:
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(""))
}
我是否需要使用适当的读写方法编写自己的Format [User]或还有其他方法?我看过 https://github.com/playframework/Play20/wiki/Scalajson但我还是有点迷茫.
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.
推荐答案
是的，编写自己的Format实例是推荐的方法.例如，给定以下类:
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(""))
}
该实例可能看起来像这样:
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)
))
}
您将像这样使用它:
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))
有关更多信息，请参见文档.
See the documentation for more information.
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