如何使用Scala将含Map的JSON转换为目标数组结构？

输入JSON #

{
    "record1": {
        "firstName": "John",
        "lastName": "Doe",
        "locations": {
            "29b2f2295cd74b8cbb53db4379f0d823": "New York"
        }
    },
    "record2": {
        "firstName": "Carol",
        "lastName": "Rees",
        "locations": {
            "0055bb74b4984156b821ebbea6937084": "California"
        }
    },
    "record3": {
        "firstName": "Colin",
        "lastName": "Scott",
        "locations": {
            "aba67f566fc24f8a8eb3165648ca5e4f": "Toronto",
            "b847750c565246638dbc72cb89ead227": "London"
        }
    }
}

目标JSON #

{
    "record1": {
        "firstName": "John",
        "lastName": "Doe",
        "locations": [{
            "id" : "29b2f2295cd74b8cbb53db4379f0d823",
            "location": "New York"
        }]
    },
    "record2": {
        "firstName": "Carol",
        "lastName": "Rees",
        "locations": [{
            "id" : "0055bb74b4984156b821ebbea6937084",
            "location": "California"
        }]
    },
    "record3": {
        "firstName": "Colin",
        "lastName": "Scott",
        "locations": [{
            "id": "aba67f566fc24f8a8eb3165648ca5e4f",
            "location": "Toronto"
        },
            {
            "id" : "b847750c565246638dbc72cb89ead227",
            "location": "London"
        }]
    }
}

完整实现代码 #

import com.fasterxml.jackson.databind.JsonMapper
import com.fasterxml.jackson.module.scala.DefaultScalaModule

// 对应输入JSON结构的case class
case class PersonEntry(firstName: String, lastName: String, locations: Map[String, String])

// 对应目标JSON结构的case class
case class LocationEntry(id: String, location: String)
case class TargetPersonEntry(firstName: String, lastName: String, locations: List[LocationEntry])

object JsonConverter {
  def main(args: Array[String]): Unit = {
    val jsonMapper = JsonMapper.builder().addModule(DefaultScalaModule).build()

    val inputJson: String = """{
        "record1": {
            "firstName": "John",
            "lastName": "Doe",
            "locations": {
                "29b2f2295cd74b8cbb53db4379f0d823": "New York"
            }
        },
        "record2": {
            "firstName": "Carol",
            "lastName": "Rees",
            "locations": {
                "0055bb74b4984156b821ebbea6937084": "California"
            }
        },
        "record3": {
            "firstName": "Colin",
            "lastName": "Scott",
            "locations": {
                "aba67f566fc24f8a8eb3165648ca5e4f": "Toronto",
                "b847750c565246638dbc72cb89ead227": "London"
            }
        }
    }"""

    // 解析输入JSON为Map[String, PersonEntry]
    val personMap: Map[String, PersonEntry] = jsonMapper.readValue(inputJson, classOf[Map[String, PersonEntry]])

    // 将输入结构转换为目标结构
    val targetMap: Map[String, TargetPersonEntry] = personMap.map { case (key, entry) =>
      val locationList = entry.locations.map { case (id, loc) => LocationEntry(id, loc) }.toList
      key -> TargetPersonEntry(entry.firstName, entry.lastName, locationList)
    }

    // 序列化为格式化后的目标JSON并输出
    val outputJson = jsonMapper.writerWithDefaultPrettyPrinter().writeValueAsString(targetMap)
    println(outputJson)
  }
}

核心步骤说明 #

1. 新增LocationEntry和TargetPersonEntry两个case class，完全匹配目标JSON的嵌套结构。
2. 遍历输入解析后的PersonEntry集合，将每个locations的键值对转换为LocationEntry实例，再转成List类型。
3. 使用Jackson的writerWithDefaultPrettyPrinter()生成带格式的JSON字符串，确保输出和目标结构一致。

内容的提问来源于stack exchange，提问作者psb