Question

我们说我有一个这样的课程：

abstract class SomeSuperClass(name: String)
case class SomeClass(someString: String, opt: Option[String]) extends SomeSuperClass("someName")

我想序列化这个类并能够添加name字段，这是我的第一个方法：

implicit def serialize: Writes[SomeClass] = new Writes[SomeClass] {
  override def writes(o: SomeClass): JsValue = Json.obj(
    "someString" -> o.someString,
    "opt" -> o.opt,
    "name" -> o.name
  )
}

如果有null，则返回None，因此我将实施following the documentation更改为：

implicit def serialize: Writes[SomeClass] = (
  (JsPath \ "someString").write[String] and
  (JsPath \ "opt").writeNullable[String] and
  (JsPath \ "name").write[String]
)(unlift(SomeClass.unapply))

这不会编译，只有在删除名称字段时才有效：

[error]   [B](f: B => (String, Option[String], String))(implicit fu: play.api.libs.functional.ContravariantFunctor[play.api.libs.json.OWrites])play.api.libs.json.OWrites[B] <and>
[error]   [B](f: (String, Option[String], String) => B)(implicit fu: play.api.libs.functional.Functor[play.api.libs.json.OWrites])play.api.libs.json.OWrites[B]
[error]  cannot be applied to (api.babylon.bridge.messaging.Command.SomeClass => (String, Option[String]))
[error]       (JsPath \ "opt").writeNullable[String] and

如何添加一个不严格存在于案例类中且具有可选字段的字段？

我正在使用play-json 2.3.0。

Answer 1

您可以为JSON写入编写自己的提取器，而不是使用默认的（编译器生成的）unapply方法，该方法对您的继承值一无所知，它采用SomeClass实例并返回(String, Option[String], String)元组，即：

implicit def serialize: Writes[SomeClass] = (
  (JsPath \ "someString").write[String] and
  (JsPath \ "opt").writeNullable[String] and
  (JsPath \ "name").write[String]
)(s => (s.someString, s.opt, s.name))

这会给你：

Json.toJson(SomeClass("foo", None))
// {"someString":"foo","name":"someName"}

Jspm.toJson(SomeClass("foo", Some("bar")))
// {"someString":"foo","opt":"bar","name":"someName"}

播放Json添加课程中不存在的字段

1 个答案: