假设:
import argonaut._
import Argonaut._
import ArgonautShapeless._
sealed trait Parent
case class Foo(x: Int) extends Parent
case class Bar(y: String) extends Parent
我试图定义DecodeJson[Parent]
:
implicit val parentDecodeJson: DecodeJson[Parent] =
DecodeJson(c => c.focus.objectFields match {
case Some("x" :: _) => implicitly[DecodeJson[Foo]].decode(c)
case _ => implicitly[DecodeJson[Bar]].decode(c)
})
然而,由于argonaut.DecodeResult
是不变的,因此失败了。
<console>:42: error: type mismatch;
found : argonaut.DecodeResult[Foo]
required: argonaut.DecodeResult[Parent]
Note: Foo <: Parent, but class DecodeResult is invariant in type A.
You may wish to define A as +A instead. (SLS 4.5)
case Some("x" :: _) => implicitly[DecodeJson[Foo]].decode(c)
^
<console>:43: error: type mismatch;
found : argonaut.DecodeResult[Bar]
required: argonaut.DecodeResult[Parent]
Note: Bar <: Parent, but class DecodeResult is invariant in type A.
You may wish to define A as +A instead. (SLS 4.5)
case _ => implicitly[DecodeJson[Bar]].decode(c)
^
所以,我想出了:
implicit val parentDecodeJson: DecodeJson[Parent] =
DecodeJson(c => c.focus.objectFields match {
case Some("x" :: _) => implicitly[DecodeJson[Foo]].decode(c).flatMap{a => DecodeResult.ok(a)}
case _ => implicitly[DecodeJson[Bar]].decode(c).flatMap{a => DecodeResult.ok(a)}
})
这似乎有效:
scala> Json.obj( ("x", jNumber(42)) ).as[Parent]
res2: argonaut.DecodeResult[Parent] = DecodeResult(Right(Foo(42)))
scala> Json.obj( ("y", jString("hi!")) ).as[Parent]
res3: argonaut.DecodeResult[Parent] = DecodeResult(Right(Bar(hi!)))
有更清洁的方式吗?
答案 0 :(得分:0)
您无需定义自己的含义:
println(EncodeJson.of[Parent].encode(Bar("a")))//{"Bar":{"y":"a"}}
println(EncodeJson.of[Parent].apply(Bar("a")))//{"Bar":{"y":"a"}}
println(Bar("a").asJson)//{"y":"a"}
println((Bar("a"): Parent).asJson)//{"Bar":{"y":"a"}}
println("{\"Bar\":{\"y\":\"a\"}}".decode[Parent])//Right(Bar(a))
println("{\"Bar\":{\"y\":\"a\"}}".decodeOption[Parent])//Some(Bar(a))
或者我是否误解了你的目标?