This is an excerpt from the Scala Cookbook (partially modified for the internet). This is a short recipe, Recipe 15.3, “How to create 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 = jValue.extract[MailServer]
The JValue class is the root class in the Lift-JSON abstract syntax tree (AST), and its extract method builds a case class instance from a JSON string.
Working with objects that have collections is a little more difficult, and that process is covered in the next recipe.