2

假设我有这样的课程：

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该name字段，这是我的第一种方法：

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
  )
}

null如果有 a 则返回None，因此我按照文档将实现更改为：

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，只有当我删除名称字段时它才有效：

[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。

4

1 回答 1

4

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，它对继承的值一无所知，它接受一个SomeClass实例并返回一个(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))

这给了你：

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

Jspm.toJson(SomeClass("foo", Some("bar")))
// {"someString":"foo","opt":"bar","name":"someName"}
于 2014-10-14T10:58:35.783 回答