Scala Cookbook

15.3. Creating a Simple Scala Object from a JSON String

Problem

You need to convert a JSON string into a simple Scala object, such as a Scala case class that has no collections.

Solution

Use the Lift-JSON library to convert a JSON string to an instance of a case class. This is referred to as deserializing the string into an object.

The following code shows a complete example of how to use Lift-JSON to convert a JSON string into a case class named MailServer. As its name implies, MailServer represents the information an email client needs to connect to a server:

import net.liftweb.json._

// a case class to represent a mail server
case class MailServer(url: String, username: String, password: String)

object JsonParsingExample extends App {

  implicit val formats = DefaultFormats

  // simulate a json string
  val jsonString = """
  {
    "url": "imap.yahoo.com",
    "username": "myusername",
    "password": "mypassword"
  }
  """

  // convert a String to a JValue object
  val jValue = parse(jsonString)

  // create a MailServer object from the string
  val mailServer = jValue.extract[MailServer]
  println(mailServer.url)
  println(mailServer.username)
  println(mailServer.password)

}

In this example, the jsonString contains the text you’d expect to receive if you called a web service asking for a MailServer instance. That string is converted into a Lift-JSON JValue object with the parse function:

val jValue = parse(jsonString)

Once you have a JValue object, use its extract method to create a MailServer object:

val mailServer ...