Converting JSON in one format to another in Scala

问题

I'm looking for suggestions or libraries that can help me convert JSON (with nested structure) from one format to another in Scala.

I saw there are a few JavaScript and Java based solutions. Anything in Scala ?

回答1:

I really like the Play JSON library. It's API is very clean and it's very fast even if some parts have a slightly steeper learning curve. You can also use the Play JSON library even if you aren't using the rest of Play.

https://playframework.com/documentation/2.3.x/ScalaJson

To convert JSON to scala objects (and vice versa), Play uses implicits. There is a Reads type which specifies how to convert JSON to a scala type, and a Writes type which specifies how to convert a scala object to JSON.

For example:

case class Foo(a: Int, b: String)

There are a few different routes you can take to convert Foo to JSON. If your object is simple (like Foo), Play JSON can create a conversion function for you:

implicit val fooReads = Json.reads[Foo]

or you can create a custom conversion function if you want more control or if your type is more complex. The below examples uses the name id for the property a in Foo:

implicit val fooReads = (
  (__ \ "id").read[Int] ~
  (__ \ "name").read[String]
)(Foo)

The Writes type has similar capabilities:

implicit val fooWrites = Json.writes[Foo]

implicit val fooWrites = (
   (JsPath \ "id").write[Int] and
   (JsPath \ "name").write[String]
)(unlift(Foo.unapply))

You can read more about Reads/Writes (and all the imports you will need) here: https://playframework.com/documentation/2.3.x/ScalaJsonCombinators

You can also transform your JSON without mapping JSON to/from scala types. This is fast and often requires less boilerplate. A simple example:

import play.api.libs.json._    

// Only take a single branch from the input json
// This transformer takes the entire JSON subtree pointed to by 
// key bar (no matter what it is)
val pickFoo = (__ \ 'foo).json.pickBranch

// Parse JSON from a string and apply the transformer
val input = """{"foo": {"id": 10, "name": "x"}, "foobar": 100}"""
val baz: JsValue = Json.parse(input)
val foo: JsValue = baz.transform(pickFoo)

You can read more about transforming JSON directly here: https://playframework.com/documentation/2.3.x/ScalaJsonTransformers

回答2:

You can use Json4s Jackson. With PlayJson, you have to write Implicit conversions for all the case classes. If the no. of classes are small, and will not have frequent changes while development, PlayJson seems to be okay. But, if the case classes are more, I recommend using json4s. You need to add implicit conversion for different types, so that json4s will understand while converting to json.

You can add the below dependency to your project to get json4s-jackson

"org.json4s" %% "json4s-jackson" % "3.2.11"

A sample code is given below (with both serialization and deserialization):

import java.util.Date
import java.text.SimpleDateFormat

import org.json4s.DefaultFormats
import org.json4s.jackson.JsonMethods._
import org.json4s.jackson.{Serialization}

/**
 * Created by krishna on 19/5/15.
 */
case class Parent(id:Long, name:String, children:List[Child])

case class Child(id:Long, name:String, simple: Simple)

case class Simple(id:Long, name:String, date:Date)

object MainClass extends App {
  implicit val formats = (new DefaultFormats {
    override def dateFormatter = new SimpleDateFormat("yyyy-MM-dd")
  }.preservingEmptyValues)
  val d = new Date()
  val simple = Simple(1L, "Simple", d)
  val child1 = Child(1L, "Child1", simple)
  val child2 = Child(2L, "Child2", simple)
  val parent = Parent(1L, "Parent", List(child1, child2))

  //Conversion from Case Class to Json
  val json = Serialization.write(parent)
  println(json)

  //Conversion from Json to Case Class
  val parentFromJson = parse(json).extract[Parent]
  println(parentFromJson)

}

来源：https://stackoverflow.com/questions/30335454/converting-json-in-one-format-to-another-in-scala

标签

json

scala

converter