How to parse JSON in Scala using standard Scala classes?

Phil picture Phil · Nov 13, 2010 · Viewed 127.1k times · Source

I am using the build in JSON class in Scala 2.8 to parse JSON code. I don't want to use the Liftweb one or any other due to minimizing dependencies.

The way I am doing it seems too imperative, is there a better way to do it?

import scala.util.parsing.json._
...
val json:Option[Any] = JSON.parseFull(jsonString)
val map:Map[String,Any] = json.get.asInstanceOf[Map[String, Any]]
val languages:List[Any] = map.get("languages").get.asInstanceOf[List[Any]]
languages.foreach( langMap => {
val language:Map[String,Any] = langMap.asInstanceOf[Map[String,Any]]
val name:String = language.get("name").get.asInstanceOf[String]
val isActive:Boolean = language.get("is_active").get.asInstanceOf[Boolean]
val completeness:Double = language.get("completeness").get.asInstanceOf[Double]
}

Answer

huynhjl picture huynhjl · Nov 15, 2010

This is a solution based on extractors which will do the class cast:

class CC[T] { def unapply(a:Any):Option[T] = Some(a.asInstanceOf[T]) }

object M extends CC[Map[String, Any]]
object L extends CC[List[Any]]
object S extends CC[String]
object D extends CC[Double]
object B extends CC[Boolean]

val jsonString =
    """
      {
        "languages": [{
            "name": "English",
            "is_active": true,
            "completeness": 2.5
        }, {
            "name": "Latin",
            "is_active": false,
            "completeness": 0.9
        }]
      }
    """.stripMargin

val result = for {
    Some(M(map)) <- List(JSON.parseFull(jsonString))
    L(languages) = map("languages")
    M(language) <- languages
    S(name) = language("name")
    B(active) = language("is_active")
    D(completeness) = language("completeness")
} yield {
    (name, active, completeness)
}

assert( result == List(("English",true,2.5), ("Latin",false,0.9)))

At the start of the for loop I artificially wrap the result in a list so that it yields a list at the end. Then in the rest of the for loop I use the fact that generators (using <-) and value definitions (using =) will make use of the unapply methods.

(Older answer edited away - check edit history if you're curious)