scala中对象与map转json

来源：互联网发布：java web项目实例 pdf 编辑：程序博客网时间：2024/05/08 07:40

今天需要把一个object对象转为jsonstring，刚开始尝试用fastjson

但是一直报

Error:(14, 18) ambiguous reference to overloaded definition,
both method toJSONString in object JSON of type (x$1: Any, x$2: com.alibaba.fastjson.serializer.SerializerFeature*)String
and method toJSONString in object JSON of type (x$1: Any)String
match argument types (Map[String,String]) and expected result type Any
println(JSON.toJSONString(maplevel1))

异常，方法重载异常，需要显示指明调用的哪个方法，

第二放弃fastjson

使用

json4s

测试代码

package project_session_spark.sparkapiimport org.json4s._import org.json4s.JsonDSL._import org.json4s.jackson.JsonMethods._/**  * Created by jevon.liu on 2017/8/16.  */object StringToJson {  case  class Person(var name:Any, var age:Any) {  }  var maplevel1:Map[String,String] = Map()  maplevel1 += ("apple"-> "苹果");  var per = Person("liu" , 2)  per.name = "sssss"  def main(args: Array[String]): Unit = {    val jsons1 = compact(render(per.toString))    println( jsons1)    val jsons2 = compact(render(maplevel1))    println(maplevel1)  }}

可以将map或者对象转为jsonstring

阅读全文

0 0