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 to

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

Tx.

Answer 1

You can do this pretty concisely with Play's JSON transformers. The following is off the top of my head, and I'm sure it could be greatly improved on:

import play.api.libs.json._

val flattenMeta = (__ \ 'metadata).read[JsObject].flatMap(
  _.fields.foldLeft((__ \ 'metadata).json.prune) {
    case (acc, (k, v)) => acc andThen __.json.update(
      Reads.of[JsObject].map(_ + (s"metadata.$k" -> v))
    )
  }
)

And then:

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

And:

scala> json.transform(flattenMeta).foreach(Json.prettyPrint _ andThen println)
{
  "metadata.id" : "1234",
  "metadata.type" : "file",
  "metadata.length" : 395
}

Just change the path if you want to handle metadata fields somewhere else in the tree.

---

Note that using a transformer may be overkill here—see e.g. Pascal Voitot's input in this thread, where he proposes the following:

(json \ "metadata").as[JsObject].fields.foldLeft(Json.obj()) {
  case (acc, (k, v)) => acc + (s"metadata.$k" -> v)
}

It's not as composable, and you'd probably not want to use as in real code, but it may be all you need.