将 Map 转换为 Scala 对象

Question

参考 [Stackoverflow] Scala: convert map to case class 我尝试复制其中一个响应，但看到以下错误：

Cannot construct instance of 'com.practice.scala.Test' (no Creators, like default construct, exist): cannot deserialize from Object value (no delegate- or property-based Creator) at [Source: UNKNOWN; line: -1, column: -1]

代码：

import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.module.scala.ScalaObjectMapper

case class Test(k1: Int, k2: String, k3: String)

object Workspace {
  def fromMap[T](map: Map[String, Any])(implicit m: Manifest[T]): T = {
    val mapper = new ObjectMapper() with ScalaObjectMapper
    mapper.convertValue(map)
  }

  def main(args: Array[String]): Unit = {
    val myMap = Map("k1" -> 1, "k2" -> "val2", "k3" -> "val3")
    val result = fromMap[Test](myMap)
    println(result)
  }
}

这个错误说明了什么？我错过了什么吗？

Answer 1

您需要在映射器上注册DefaultScalaModule：

// With 2.10 and later
val mapper = JsonMapper.builder()
  .addModule(DefaultScalaModule)
  .build()

// versions before 2.10 (also support for later 2.x but not 3.0)
val mapper = new ObjectMapper()
mapper.registerModule(DefaultScalaModule)

来源：https://github.com/FasterXML/jackson-module-scala#usage