scala map转化为json string

2019独角兽企业重金招聘Python工程师标准>>>

map数据

val m = Map(
    "name" -> "john doe", 
    "age" -> 18, 
    "hasChild" -> true, 
    "childs" -> List(
        Map("name" -> "dorothy", "age" -> 5, "hasChild" -> false),
        Map("name" -> "bill", "age" -> 8, "hasChild" -> false)
    )
)

期望String数据

{
    "name": "john doe",
    "age": 18,
    "hasChild": true,
    "childs": [
        {
            "name": "dorothy",
            "age": 5,
            "hasChild": false
        },
        {
            "name": "bill",
            "age": 8,
            "hasChild": false
        }
    ]
}

方法一

import org.json4s._
import org.json4s.native.Serialization._
import org.json4s.native.Serialization
implicit val formats = Serialization.formats(NoTypeHints)

 val m = Map(
  "name" -> "john doe",
  "age" -> 18,
  "hasChild" -> true,
  "childs" -> List(
    Map("name" -> "dorothy", "age" -> 5, "hasChild" -> false),
    Map("name" -> "bill", "age" -> 8, "hasChild" -> false)))

 write(m)

方法二

import org.json4s.native.Json
import org.json4s.DefaultFormats

Json(DefaultFormats).write(m)

方法三

myString.substring(1, myString.length - 1)
        .split(",")
        .map(_.split(":"))
        .map { case Array(k, v) => (k.substring(1, k.length-1), v.substring(1, v.length-1))}
        .toMap

json string转化为map

val m = Map(
      "name" -> "john doe",
      "age" -> 18,
      "hasChild" -> true,
      "childs" -> List(
        Map("name" -> "dorothy", "age" -> 5, "hasChild" -> false),
        Map("name" -> "bill", "age" -> 8, "hasChild" -> false)))

val a = m.values.asInstanceOf[Map[String,Any]]

 

转载于:https://my.oschina.net/u/2000675/blog/1818718