24

有两个类FooBarFoo包含一个字段Bar。问题是,如何Writes为 class实现隐式 json Foo

这是代码:

package models

import play.api.libs.json._

case class Foo(id: String, bar: Bar)

object Foo {
  implicit val implicitFooWrites = new Writes[Foo] {
    def writes(foo: Foo): JsValue = {
      Json.obj(
        "id" -> foo.id,
        "bar" -> foo.bar
      )
    }
  }
}

case class Bar(x: String, y: Int)

object Bar {
  implicit val implicitBarWrites = new Writes[Bar] {
    def writes(bar: Bar): JsValue = {
      Json.obj(
        "x" -> bar.x,
        "y" -> bar.y
      )
    }
  }
}

当我尝试编译时,出现以下错误:

没有为类型 models.Bar 找到 Json 反序列化器。尝试为此类型实现隐式写入或格式。

我不明白这个编译器错误,因为我为 models.Bar 类实现了隐式写入。这里有什么问题?

4

1 回答 1

33

这是一个可见性问题,在声明隐式 Writes[Foo] 时,您不会使其隐式 Writes[Bar] 可见:

scala> :paste
// Entering paste mode (ctrl-D to finish)

import play.api.libs.json._

case class Bar(x: String, y: Int)

object Bar {
  implicit val implicitBarWrites = new Writes[Bar] {
    def writes(bar: Bar): JsValue = {
      Json.obj(
        "x" -> bar.x,
        "y" -> bar.y
      )
    }
  }
}

case class Foo(id: String, bar: Bar)

object Foo {

  import Bar._

  implicit val implicitFooWrites = new Writes[Foo] {
    def writes(foo: Foo): JsValue = {
      Json.obj(
        "id" -> foo.id,
        "bar" -> foo.bar
      ) 
    } 
  }     
}

// Exiting paste mode, now interpreting.

import play.api.libs.json._
defined class Bar
defined module Bar
defined class Foo
defined module Foo

scala> Json.prettyPrint(Json.toJson(Foo("23", Bar("x", 1))))
res0: String = 
{
  "id" : "23",
  "bar" : {
    "x" : "x",
    "y" : 1
  }
}

此外,如果您使用的是 Play 2.1+,请务必查看 2.10 宏的全新用法:http ://www.playframework.com/documentation/2.1.0/ScalaJsonInception

如果您对案例类的使用感到满意,并且 val/vars 的名称被用作 json 输出中的键,就像您的案例 BTW 一样,那么您可以使用两个单行:

implicit val barFormat = Json.writes[Bar]
implicit val fooFormat = Json.writes[Foo]

这将为您提供确切的等价物:

scala> import play.api.libs.json._
import play.api.libs.json._

scala> case class Bar(x: String, y: Int)
defined class Bar

scala> case class Foo(id: String, bar: Bar)
defined class Foo

scala> implicit val barWrites = Json.writes[Bar]
barWrites: play.api.libs.json.OWrites[Bar] = play.api.libs.json.OWrites$$anon$2@257cae95

scala> implicit val fooWrites = Json.writes[Foo]
fooWrites: play.api.libs.json.OWrites[Foo] = play.api.libs.json.OWrites$$anon$2@48f97e2a

scala> Json.prettyPrint(Json.toJson(Foo("23", Bar("x", 1))))
res0: String = 
{
  "id" : "23",
  "bar" : {
    "x" : "x",
    "y" : 1
  }
}
于 2013-03-29T14:45:32.560 回答