Play [Scala]: How to flatten a JSON object

Question

Given the following JSON...

{
  \"metadata\": {
    \"id\": \"1234\",
    \"type\": \"file\",
    \"length\": 395
  }
}

... how do I convert it

Answer 1

This is definitely not trivial, but possible by trying to flatten it recursively. I haven't tested this thoroughly, but it works with your example and some other basic one's I've come up with using arrays:

object JsFlattener {

    def apply(js: JsValue): JsValue = flatten(js).foldLeft(JsObject(Nil))(_++_.as[JsObject])

    def flatten(js: JsValue, prefix: String = ""): Seq[JsValue] = {
        js.as[JsObject].fieldSet.toSeq.flatMap{ case (key, values) =>
            values match {
                case JsBoolean(x) => Seq(Json.obj(concat(prefix, key) -> x))
                case JsNumber(x) => Seq(Json.obj(concat(prefix, key) -> x))
                case JsString(x) => Seq(Json.obj(concat(prefix, key) -> x))
                case JsArray(seq) => seq.zipWithIndex.flatMap{ case (x, i) => flatten(x, concat(prefix, key + s"[$i]")) }  
                case x: JsObject => flatten(x, concat(prefix, key))
                case _ => Seq(Json.obj(concat(prefix, key) -> JsNull))
            }
        }
    }

    def concat(prefix: String, key: String): String = if(prefix.nonEmpty) s"$prefix.$key" else key

}

JsObject has the fieldSet method that returns a Set[(String, JsValue)], which I mapped, matched against the JsValue subclass, and continued consuming recursively from there.

You can use this example by passing a JsValue to apply:

val json = Json.parse("""
    {
      "metadata": {
        "id": "1234",
        "type": "file",
        "length": 395
      }
    }
"""
JsFlattener(json)

We'll leave it as an exercise to the reader to make the code more beautiful looking.