What's the simplest way to parse JSON in Scala?

While XML is a first-class citizen in Scala, there’s no “default” way to parse JSON. So searching StackOverflow and Google yields all kinds of responses that seem unnecessarily complicated.

Jackson

This SO answer describes the easiest solution, which gives you a Map[String, Object], use jackson-module-scala.

import scala.io._
import com.fasterxml.jackson.databind.ObjectMapper
import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper

object Main {
  def main(args: Array[String]): Unit = {
    val filename = args.head
    // read
    println(s"Reading ${args.head} ...")
    val json = Source.fromFile(filename)
    // parse
    val mapper = new ObjectMapper() with ScalaObjectMapper
    mapper.registerModule(DefaultScalaModule)
    val parsedJson = mapper.readValue[Map[String, Object]](json.reader())
    println(parsedJson)
  }
}

Here’s the output from a sample JSON.

Reading example.json ...
Map(glossary -> Map(title -> example glossary, GlossDiv -> ..., GlossTerm -> Standard Generalized Markup Language)))))

Liftweb JSON

The lift-json JSON parser in Liftweb does a good job, too, but returns JObject-like types instead of raw String or Map[String, Object].

import scala.io._
import net.liftweb.json._

object Main {
  def main(args: Array[String]): Unit = {
    val filename = args.head
    // read
    println(s"Reading ${args.head} ...")
    val json = Source.fromFile(filename)
    // parse
    let parsedJson = net.liftweb.json.parse(json.mkString)
    println(parsedJson)
  }
}

Here’s the output from a sample JSON.

JObject(List(JField(glossary, ...)))

Source

The code above is here, maybe someone can contribute an example for writing JSON?

Daniel Doubrovkine

What's the simplest way to parse JSON in Scala?

Jackson

Liftweb JSON

Source