Scala3下替换已废弃ScalaObjectMapper实现JSON序列化反序列化

问题说明 #

原有基于ScalaObjectMapper实现的JSON序列化/反序列化工具类出现废弃告警：

ScalaObjectMapper is deprecated because Manifests are not supported in Scala3
官方指定替代方案为使用ClassTagExtensions，改造后可完全保留原有使用习惯，无需修改业务侧调用代码。

核心改造点 #

• 导入替换：删除com.fasterxml.jackson.module.scala.ScalaObjectMapper导入，新增com.fasterxml.jackson.module.scala.ClassTagExtensions导入
• Mapper定义修改：将new ObjectMapper() with ScalaObjectMapper改为new ObjectMapper() with ClassTagExtensions
• 隐式参数替换：所有泛型方法的隐式参数从Manifest[T]替换为scala.reflect.ClassTag[T]
• 序列化逻辑无需调整，和原有实现完全一致

改造后完整代码 #

import com.fasterxml.jackson.databind.{DeserializationFeature, ObjectMapper}
import com.fasterxml.jackson.module.scala.ClassTagExtensions
import com.fasterxml.jackson.module.scala.DefaultScalaModule
import scala.reflect.ClassTag

object JsonUtil {
  val mapper = new ObjectMapper() with ClassTagExtensions
  mapper.registerModule(DefaultScalaModule)
  mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)

  def toJson(value: Map[Symbol, Any]): String = {
    toJson(value map { case (k, v) => k.name -> v })
  }

  def toJson(value: Any): String = {
    mapper.writeValueAsString(value)
  }

  def toMap[V](json: String)(implicit m: ClassTag[V]): Map[String, V] = fromJson[Map[String, V]](json)

  def fromJson[T](json: String)(implicit m: ClassTag[T]): T = {
    mapper.readValue[T](json)
  }
}

object MarshallableImplicits {

  implicit class Unmarshallable(unMarshallMe: String) {
    def toMap: Map[String,Any] = JsonUtil.toMap(unMarshallMe)
    def toMapOf[V]()(implicit m: ClassTag[V]): Map[String,V] = JsonUtil.toMap[V](unMarshallMe)
    def fromJson[T]()(implicit m: ClassTag[T]): T =  JsonUtil.fromJson[T](unMarshallMe)
  }

  implicit class Marshallable[T](marshallMe: T) {
    def toJson: String = JsonUtil.toJson(marshallMe)
  }
}

用法说明 #

原有业务侧调用代码完全不需要修改，和改造前使用方式一致：

反序列化用法 #

import com.some.package.json.JsonUtil.fromJson
import com.some.package.json.MarshallableImplicits._
def deserialize(json: String): MyCaseClass = fromJson[MyCaseClass](json)

序列化用法 #

import com.some.package.json.MarshallableImplicits._
def serialize(obj: MyCaseClass): String = obj.toJson

内容的提问来源于stack exchange，提问作者radumanolescu