我需要一个简单的JSON序列化解决方案与最小的仪式。所以我很高兴找到这个即将到来的Play 2.2库。这对于普通大小写类非常有效,例如
import play.api.libs.json._
sealed trait Foo
case class Bar(i: Int) extends Foo
case class Baz(f: Float) extends Foo
implicit val barFmt = Json.format[Bar]
implicit val bazFmt = Json.format[Baz]
但是以下操作失败:
implicit val fooFmt = Json.format[Foo] // "No unapply function found"
我如何为Foo
设置所谓的缺失提取器?
修改2015-09-22
库play-json-extra包括play-json-variant策略,但也包括[play-json-extensions]策略(用于case对象的平面字符串与case类的对象混合,除非需要,否则没有额外的$variant或$type)。它还为基于宏格式的枚举提供了序列化器和反序列化器。
之前回答现在有一个叫做play-json- variables的库,它允许你这样写:
implicit val format: Format[Foo] = Variants.format[Foo]
这将自动生成相应的格式,它还将通过添加$variant属性(相当于0__的class
属性)来处理以下情况的消歧
sealed trait Foo
case class Bar(x: Int) extends Foo
case class Baz(s: String) extends Foo
case class Bah(s: String) extends Foo
将生成
val bahJson = Json.obj("s" -> "hello", "$variant" -> "Bah") // This is a `Bah`
val bazJson = Json.obj("s" -> "bye", "$variant" -> "Baz") // This is a `Baz`
val barJson = Json.obj("x" -> "42", "$variant" -> "Bar") // And this is a `Bar`
下面是Foo
伴侣对象的手动实现:
implicit val barFmt = Json.format[Bar]
implicit val bazFmt = Json.format[Baz]
object Foo {
def unapply(foo: Foo): Option[(String, JsValue)] = {
val (prod: Product, sub) = foo match {
case b: Bar => (b, Json.toJson(b)(barFmt))
case b: Baz => (b, Json.toJson(b)(bazFmt))
}
Some(prod.productPrefix -> sub)
}
def apply(`class`: String, data: JsValue): Foo = {
(`class` match {
case "Bar" => Json.fromJson[Bar](data)(barFmt)
case "Baz" => Json.fromJson[Baz](data)(bazFmt)
}).get
}
}
sealed trait Foo
case class Bar(i: Int ) extends Foo
case class Baz(f: Float) extends Foo
implicit val fooFmt = Json.format[Foo] // ça marche!
验证:
val in: Foo = Bar(33)
val js = Json.toJson(in)
println(Json.prettyPrint(js))
val out = Json.fromJson[Foo](js).getOrElse(sys.error("Oh no!"))
assert(in == out)
或者直接的格式定义:
implicit val fooFmt: Format[Foo] = new Format[Foo] {
def reads(json: JsValue): JsResult[Foo] = json match {
case JsObject(Seq(("class", JsString(name)), ("data", data))) =>
name match {
case "Bar" => Json.fromJson[Bar](data)(barFmt)
case "Baz" => Json.fromJson[Baz](data)(bazFmt)
case _ => JsError(s"Unknown class '$name'")
}
case _ => JsError(s"Unexpected JSON value $json")
}
def writes(foo: Foo): JsValue = {
val (prod: Product, sub) = foo match {
case b: Bar => (b, Json.toJson(b)(barFmt))
case b: Baz => (b, Json.toJson(b)(bazFmt))
}
JsObject(Seq("class" -> JsString(prod.productPrefix), "data" -> sub))
}
}
现在理想情况下,我想自动生成apply
和unapply
方法。看来我将需要使用反射或潜入宏
Play 2.7
play-json
中支持sealed traits
object Foo{
implicit val format = Json.format[Foo]
}
玩2.6
现在可以使用play-json-derived-codecs轻松地完成此操作
就加上这个:
object Foo{
implicit val jsonFormat: OFormat[Foo] = derived.oformat[Foo]()
}
完整示例见这里:ScalaFiddle
对于0__关于直接格式定义的上一个答案的一个小修复- read方法不起作用,这里是我对它的重构,也变得更习惯-
def reads(json: JsValue): JsResult[Foo] = {
def from(name: String, data: JsObject): JsResult[Foo] = name match {
case "Bar" => Json.fromJson[Bar](data)(barFmt)
case "Baz" => Json.fromJson[Baz](data)(bazFmt)
case _ => JsError(s"Unknown class '$name'")
}
for {
name <- (json "class").validate[String]
data <- (json "data").validate[JsObject]
result <- from(name, data)
} yield result
}