How to remove key but keep childs

I'm new to Scala and I'm trying to process json document. I'm using scala 2.13.3 with the following librairies :

libraryDependencies += "org.scalatest" %% "scalatest" % "3.2.0" % "test"
libraryDependencies += "org.json4s" %% "json4s-native" % "3.6.9"

The problem is that I can't find a way to remove a key and still keep his children as follows:

Here's the json document to begin with:

{
  "Key": {
    "opt": {
      "attribute1": 0,
      "attribute2": 1,
      "attribute3": 2
    },
    "args": {
      "arg1": 0,
      "arg2": 1,
      "arg3": 2
    }
  }
}

I would like to remove the "Key" to keep only his children "opt" and "args" so that I get the Json document below :

{
  "opt": {
    "attribute1": 0,
    "attribute2": 1,
    "attribute3": 2
  },
  "args": {
    "arg1": 0,
    "arg2": 1,
    "arg3": 2
  }
}

My code

I use the Json4s library to manipulate documents, there is a transformField operator that allows to perform operations on fields (key, value) => (key, value). So I tried to define an empty key "" but it doesn't answer my need. I also tried to return only the associated value but the partial function doesn't allow it.

Here is my scala code :

val json: JObject =
  "Key" -> (
    "opt" -> (
      ("attribute1" -> 0) ~
        ("attribute2" -> 1) ~
        ("attribute3" -> 2)
      ) ~
      ("args" -> (
        ("arg1", 0) ~
          ("arg2", 1) ~
          ("arg3", 2)
        )))

val res = json transformField {
  case JField("Key", attr) => attr
}

println(pretty(render(res)))

unfortunatly I can't just use transformField to transform ("Key", attr) into attr.

Is there an easy way to remove the "Key" key from my Json while keeping its children "opt" and "args"?

Solution

It is usually better to convert JSON to Scala objects, manipulate the Scala objects, and then convert back to JSON.

This is what it might look like using jackson:

import org.json4s.{DefaultFormats, Extraction}
import org.json4s.jackson.{JsonMethods, Serialization}

val jsonIn = """
  {
    "Key": {
      "opt": {
        "attribute1": 0,
        "attribute2": 1,
        "attribute3": 2
      },
      "args": {
        "arg1": 0,
        "arg2": 1,
       "arg3": 2
     }
   }
 }
"""

case class Opt(attribute1: Int, attribute2: Int, attribute3: Int)
case class Args(arg1: Int, arg2: Int, arg3: Int)
case class Key(opt: Opt, args: Args)
case class DataIn(Key: Key)

implicit val formats = DefaultFormats

val dataIn: DataIn = Extraction.extract[DataIn](JsonMethods.parse(jsonIn))

val jsonOut: String = Serialization.write(dataIn.Key)

In this case the Scala processing is just extracting the Key field from the DataIn class.

I am on Scala 2.12 so YMMV.