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How to convert json data into case class using scala

Contributor

Hi,

In my case i need to get feed from json and that need to be streamed via spark - scala, so i need to know how to convert json data into case class using scala.

For example: {"Id":"1","Name":"Babu","Address":"dsfjskkjfs"} to (1,babu,dsfjskkjfs}

1 REPLY 1

Re: How to convert json data into case class using scala

There are several Scala native JSON libraries available (json4s, play-json, circe, as well as others) so there are lots of ways to do this.

With play-json, your solution might look like:

import play.api.libs.json._
import play.api.libs.json.Reads._
import play.api.libs.functional.syntax._

val jsonStr = """{"Id":1,"Name":"Babu","Address":"dsfjskkjfs"}"""

Person(id: Int, name: String, address: String)

implicit val personReads: Reads[Person] = (
  (JsPath \ "Id").read[Int] and
  (JsPath \ "Name").read[String] and
  (JsPath \ "Address").read[String]
)(Person.apply _)

jsonStr.validate[Person] match {
  case JsSuccess(person, _) => {
    // do something
  }
  case JsError(e) => {
    // do something
  }
}

With circe, your solution might look like:

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

val jsonStr = """{"Id":1,"Name":"Babu","Address":"dsfjskkjfs"}"""

Person(id: Int, name: String, address: String)

decode[Person](jsonStr) match {
  case Right(person) => {
    // do something with Person(1, "Babu", "dsfjskkjfs")
  }
  case Left(e) => {
    // handle error
  }
}

In your question you asked for a case class but wrote it out as a tuple, so just to close that loop, if you want to subsequently convert the data to a tuple you can do `Person.unapply(person).get`.