CODEWITHSUNDEEP
jsonscalajerkson
mane
mane

Reputation: 1179

Parsing JSON with multiple tuples to List<object> in scala

[
{"fname":"Foo","lname":"Pacman"},
{"fname":"Bar","lname":"Mario"},
{"fname":"Poo","lname":"Wario"}
]

Well I have JSON string in this format, Now what I need is to convert each tuples -> {"fname":"Foo","lname":"Pacman"}

To a Person object, for e.g. lets assume I have a case class

case class Person(fname:String,lname:String)

Now how am I to get, List<person>

If I had a JSON containing data for single tuple, then I could,

val o:Person = parse[Person](jsonString)// I am actually using Jerkson Lib

But since there are more than one tuples, how am i to parse them individually and create objects and create a list.

Upvotes: 0

Views: 1810

Answers (2)

Ashalynd
Ashalynd

Reputation: 12573

You can use json4s (which is a wrapper around either jackson or lift-json) where you also get such parsing capabilities out of the box.

   import org.json4s._
   import org.json4s.jackson.JsonMethods._
   implicit val formats = DefaultFormats 

    val personJson = """
      [
      {"fname":"Foo","lname":"Pacman"},
      {"fname":"Bar","lname":"Mario"},
      {"fname":"Poo","lname":"Wario"}
      ]"""
    case class Person(fname:String,lname:String)
    val people = parse(personJson).extract[List[Person]]

Upvotes: 2

cmbaxter
cmbaxter

Reputation: 35463

Jerkson supports deserializing lists of objects out of the box, so all you should need to do is:

val people = parse[List[Person]](personJson)

Upvotes: 2

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