如何在Scala中使用Json4s序列化密封的抽象类? [英] How to serialize sealed abstract class with Json4s in Scala?

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问题描述

如何在Scala中使用Json4s序列化密封的抽象类?

How do i serialize a sealed abstract class with Json4s in Scala?

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:

I create an object of type Customer:

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

然后我序列化为JSON:

Then I serialize to JSON:

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

那很好.但随后我尝试反序列化:

That works fine. But then I try to deserialize:

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

在这里我总是收到错误消息:

Here I keep getting error:

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

org.json4s.package$MappingException: Parsed JSON values do not match with class constructor

我花了很多时间试图使这项工作...

I spent a lot of time trying to make this work...

推荐答案

经过大量的搜索并仔细阅读了Json4s文档后,我发现我需要一个自定义的Serializer才能使其正常工作.

After a lot of googling around and a close read of the Json4s documentation i found out that I need a custom Serializer to make it work.

但是,我花了一段时间才弄清楚我需要设置格式实际使用自定义序列化器.

However, it took me a while to figure out that I need to set the formats to actually use the custom Serializer.

这是对我有用的代码.

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)

自定义序列化器:

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

scala> val serialized = write(customer)

scala> val serialized = write(customer)

序列化:字符串= {名称":"Joe",年龄":35,配偶":{名称":"Marilyn",年龄":33}}

serialized: String = {"name":"Joe","age":35,"spouse":{"name":"Marilyn","age":33}}

scala> val parsedObj = Serialization.readPerson

scala> val parsedObj = Serialization.readPerson

parsedObj:人员=客户(Joe,35,Spouse(Marilyn,33))

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

这篇关于如何在Scala中使用Json4s序列化密封的抽象类?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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