For example, I have this Map value in Scala:
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)
)
)
I want to convert it to its JSON string representation:
{
"name": "john doe",
"age": 18,
"hasChild": true,
"childs": [
{
"name": "dorothy",
"age": 5,
"hasChild": false
},
{
"name": "bill",
"age": 8,
"hasChild": false
}
]
}
I'm currenly working on Play framework v2.3, but the solution doesn't need to use Play JSON library, although it will be nice if someone can provide both Play and non-Play solution.
This is what I have done so far without success:
// using jackson library
val mapper = new ObjectMapper()
val res = mapper.writeValueAsString(m)
println(res)
Result:
{"empty":false,"traversableAgain":true}
I don't understand why I got that result.
8 Answers
As a non play solution, you can consider using json4s which provides a wrapper around jackson and its easy to use. If you are using json4s then you can convert map to json just by using:
write(m)
//> res0: String = {"name":"john doe","age":18,"hasChild":true,"childs":[{"name":"dorothy","age":5,"hasChild":false},{"name":"bill","age":8,"hasChild":false}]}
--Updating to include the full example--
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)
Output:
res0: String = {"name":"john doe","age":18,"hasChild":true,"childs":[{"name"
:"dorothy","age":5,"hasChild":false},{"name":"bill","age":8,"hasChild":false }]}
Alternative way:
import org.json4s.native.Json
import org.json4s.DefaultFormats
Json(DefaultFormats).write(m)
val mapper = new ObjectMapper()
mapper.writeValueAsString(Map("a" -> 1))
result> {"empty":false,"traversableAgain":true}
==============================
import com.fasterxml.jackson.module.scala.DefaultScalaModule
val mapper = new ObjectMapper()
mapper.registerModule(DefaultScalaModule)
mapper.writeValueAsString(Map("a" -> 1))
result> {"a":1}
You need to tell jackson how to deal with scala objects: mapper.registerModule(DefaultScalaModule)
val mymap = array.map {
case 1 => ("A", 1)
case 2 => ("B", 2)
case 3 => ("C", 3)
}
.toMap
Using scala.util.parsing.json.JSONObject, you only need 1 line:
import scala.util.parsing.json.JSONObject
JSONObject(mymap).toString()
If you're working with a well-defined data model, why not define case classes and use Play JSON macros to handle conversion? i.e.
case class Person(name: String, age: Int, hasChild: Boolean, childs: List[Person])
implicit val fmt = Json.format[Person]
val person = Person(...)
val jsonStr = Json.toJson(person)
One thing you can do using the Jackson library is to use a java HashMap object, instead of a Scala one. Then you can basically use the same "without success" code you already wrote.
import org.codehaus.jackson.map.ObjectMapper
val mapper = new ObjectMapper()
val jmap = new java.util.HashMap[String, Int]()
jmap.put("dog", 4)
jmap.put("cat", 1)
// convert to json formatted string
val jstring = mapper.writeValueAsString(jmap)
println(jstring)
returns
jstring: String = {"dog":4,"cat":1}
In case somebody is looking for a solution using standard libraries.
def toJson(query: Any): String = query match {
case m: Map[String, Any] => s"{${m.map(toJson(_)).mkString(",")}}"
case t: (String, Any) => s""""${t._1}":${toJson(t._2)}"""
case ss: Seq[Any] => s"""[${ss.map(toJson(_)).mkString(",")}]"""
case s: String => s""""$s""""
case null => "null"
case _ => query.toString
}
The following snippet extracted from the official scala site worked well for me. It uses uPickle library.
val map: Map[String, Int] =
Map("Toolkitty" -> 3, "Scaniel" -> 5)
val jsonString: String = upickle.default.write(map)
println(jsonString)
// prints: {"Toolkitty":3,"Scaniel":5}
