Question

如何在Scala中使用Json4s序列化密封的抽象类？

定义了以下类：

sealed abstract class Person extends Product with Serializable
case class Spouse(name: String, age: Int) extends Person
case class Customer(name: String, age: Int, spouse: Spouse) extends Person

我创建了一个Customer类型的对象：

val customer: Customer = Customer("Joe", 35, Spouse("Marilyn", 33))

然后我序列化为JSON：

implicit val formats = DefaultFormats
val serialized = write(customer)

工作正常。但后来我尝试反序列化：

val parsedObj = Serialization.read[Person](serialized)

我不断收到错误：

org.json4s.package $ MappingException：解析的JSON值与类构造函数不匹配

我花了很多时间试图做这项工作......

Answer 1

经过大量的谷歌搜索和仔细阅读Json4s文档后，我发现我需要一个自定义的串行器来使其工作。

但是，我花了一些时间才弄清I need to set the formats实际使用自定义序列化程序。

这是适用于我的代码。

简单示例：

import org.json4s._
import org.json4s.native.JsonMethods._
import org.json4s.native.Serialization.{read, write}
import org.json4s.native.Serialization

sealed abstract class Person extends Product with Serializable
case class Spouse(name: String, age: Int) extends Person
case class Customer(name: String, age: Int, spouse: Spouse) extends Person

val customer: Customer = Customer("Joe", 35, Spouse("Marilyn", 33))

implicit val formats = Serialization.formats(NoTypeHints) + PersonSerializer
val serialized = write(customer)    
val parsedObj = Serialization.read[Person](serialized)

自定义序列化程序：

object PersonSerializer extends Serializer[Person] {
    private val PersonClass = classOf[Person]

    def deserialize(implicit format: Formats)
    : PartialFunction[(TypeInfo, JValue), Person] = {
        case (TypeInfo(PersonClass, _), json) =>
            json match {
                case JObject(List(
                    JField("name", JString(d)),
                    JField("age", JInt(f)),
                    ("spouse", JObject(List(JField("name", JString(g)), JField("age", JInt(h)))))
                    )) => Customer(d, f.toInt, Spouse(g, h.toInt))
                case JObject(List(
                    JField("name", JString(d)),
                    JField("age", JInt(f))
                    )) => Spouse(d, f.toInt)
            }
    }

    def serialize(implicit format: Formats): PartialFunction[Any, JValue] = {
        case x: Customer =>
            JObject(List(
                JField("name", JString(x.name)),
                JField("age", JInt(x.age)),
                ("spouse", JObject(List(JField("name", JString(x.spouse.name)), JField("age", JInt(x.spouse.age)))))
            ))
        case x: Spouse =>
            JObject(List(
                JField("name", JString(x.name)),
                JField("age", JInt(x.age))
            ))
    }
}

输出

阶＆GT; val serialized = write（customer）

序列化：字符串= {“姓名”：“乔”，“年龄”：35，“配偶”：{“姓名”：“玛丽莲”，“年龄”：33}}

阶＆GT; val parsedObj = Serialization.readPerson

parsedObj：Person = Customer（Joe，35，Spouse（Marilyn，33））

如何在Scala中使用Json4s序列化密封的抽象类？

1 个答案: