What is the most straightforward way to parse JSON in Scala?

问题

I'm working on a simple web application with Scala. The plan is to obtain JSON data from an external API, and insert it into a template (unfortunately, obtaining the data in XML is not an option).

I've tried working with Twitter's scala-json library, but I can't get it to compile properly (the code on github fails to update in sbt, saying standard-project 7.10 is not available and I haven't worked that out yet).

lift-json looks impressive, but appears to be a lot more elaborate than I need right now.

Trying to import a library I've worked with in Java, jsonic, results in various arcane errors. This is too bad because I rather like how straightforward jsonic is.

I've made a bit of progress with the built in scala.util.parsing.json.JSON, but actually I can't tell how to access the elements. I'm somewhat new to Scala, as you may have noted. How do you access the properties of JSONObjects?

scala.util.parsing.json.JSON has a lot of information, but is there a straightforward tutorial on how to use this anywhere?

I'm really only interested in deserializing JSON at the moment, to Ints, Strings, Maps and Lists. I don't have a need to serialize objects or make the deserialized objects fit into a class at the moment.

Can anyone point me to ways to work with one of the aforementioned libraries, or help me get set up with a Java lib that will do what I want?

回答1:

Lift JSON provides several different styles of deserializing JSON. Each have their pros and cons.

val json = JsonParser.parse(""" { "foo": { "bar": 10 }} """)

LINQ style query comprehension:

scala> for { JField("bar", JInt(x)) <- json } yield x 

res0: List[BigInt] = List(10)

More examples: http://github.com/lift/lift/blob/master/framework/lift-base/lift-json/src/test/scala/net/liftweb/json/QueryExamples.scala

Extract values with case classes

implicit val formats = net.liftweb.json.DefaultFormats 
case class Foo(foo: Bar) 
case class Bar(bar: Int) 
json.extract[Foo] 

More examples: https://github.com/lift/lift/blob/master/framework/lift-base/lift-json/src/test/scala/net/liftweb/json/ExtractionExamples.scala

XPath style

scala> val JInt(x) = json \ "foo" \ "bar"

x: BigInt = 10

Non-type safe values

scala> json.values

res0: Map((foo,Map(bar -> 10)))

回答2:

Here are quick examples of deserialising raw string JSON into case class model for different Scala JSON libraries:

play-json

import play.api.libs.json._

case class User(id: Int, name: String)

object User {
  implicit val codec = Json.format[User]
}

object PlayJson extends App {
  val string = """{"id": 124, "name": "John"}"""
  val json = Json.parse(string)
  val user = json.as[User]
  println(user)
}

lift-json

import net.liftweb.json._

case class User(id: Int, name: String)

object LiftJson extends App {
  implicit val codec = DefaultFormats
  val string = """{"id": 124, "name": "John"}"""
  val json = parse(string)
  val user = json.extract[User]
  println(user)
}

spray-json

import spray.json._
import DefaultJsonProtocol._

case class User(id: Int, name: String)

object UserJsonProtocol extends DefaultJsonProtocol {
  implicit val codec = jsonFormat2(User)
}

object SprayJson extends App {
  import UserJsonProtocol._
  val string = """{"id": 124, "name": "John"}"""
  val json = string.parseJson
  val user = json.convertTo[User]
  println(user)
}

sphere-json

import io.sphere.json.generic._
import io.sphere.json._

case class User(id: Int, name: String)

object SphereJson extends App {
  implicit val codec = deriveJSON[User]
  val string = """{"id": 124, "name": "John"}"""
  val user = fromJSON[User](string)
  println(user)
}

argonaut

import argonaut._ 
import Argonaut._

case class User(id: Int, name: String)

object ArgonautJson extends App {
  implicit def codec = casecodec2(User.apply, User.unapply)("id", "name")
  val string = """{"id": 124, "name": "John"}"""
  val user = string.decodeOption[User]
  println(user)
}

circe

import io.circe.generic.auto._
import io.circe.parser._

case class User(id: Int, name: String)

object CirceJson extends App {
  val string = """{"id": 124, "name": "John"}"""
  val user = decode[User](string)
  println(user)
}

Here are dependencies for above examples:

resolvers += Resolver.bintrayRepo("commercetools", "maven")

libraryDependencies ++= Seq(

  "com.typesafe.play" %% "play-json"      % "2.6.7",
  "net.liftweb"       %% "lift-json"      % "3.1.1",
  "io.spray"          %% "spray-json"     % "1.3.3",
  "io.sphere"         %% "sphere-json"    % "0.9.0",
  "io.argonaut"       %% "argonaut"       % "6.2",
  "io.circe"          %% "circe-core"     % "0.8.0",
  "io.circe"          %% "circe-generic"  % "0.8.0",
  "io.circe"          %% "circe-parser"   % "0.8.0"
)

This post was inspired by the following article: A quick tour of JSON libraries in Scala

Related SO question: What JSON library to use in Scala?

来源：https://stackoverflow.com/questions/4169153/what-is-the-most-straightforward-way-to-parse-json-in-scala