Support for untyped or "metadata" field about scala-jsonschema HOT 4 CLOSED

hi, Grant
lets start from, how do you want your metadata object to be described in json-schema terms?
can you give me an example?

from scala-jsonschema.

something like this?

{
  "type": "object",
  "additionalProperties": true
}

from scala-jsonschema.

I want to create a field in my schema named "metadata" that has a type of "object". I need the scala type of that field to be play.api.libs.json.JsObject because that allows me to create a play-json Format which will understand how to parse the field properly.

So:

case class Payload(id: String, name: String, metadata: JsObject)
implicit val payloadFmt = Json.format[Payload]
val personSchema: json.Schema[Person] = Json.schema[Person]

should compile and

println(JsonFormatter.format(AsValue.schema(personSchema))

Should print something like:

{
  "$schema": "http://json-schema.org/draft-04/schema#",
  "type": "object",
  "additionalProperties": false,
  "properties": {
    "id": {
      "type": "string"
    },
    "name": {
      "type": "string"
    },
    "payload": {
      "type": "object"
    }
  }
}

from scala-jsonschema.

hey @gavares
please find this in 0.5.0
it won't happen automagically, though
some documentation you can find in here https://github.com/andyglow/scala-jsonschema#free-objects
please also take a look at

from scala-jsonschema.