将JSON序列化为Scala case类

Question

I need to serialize JSON to scala case class. It is not a question about JSON serialization, but is a question about class mapping in scala.

The example of JSON:

{
  "id": 98433,
  "name": "Santa Cruz Bronson",
  "vendor": {
    "id": 344,
    "name": "Santa Cruz"
  },
  "category": {
    "id": 132,
    "name": "Bicycles"
  },
  "annotation": "The best downhill cycle",
  "description": "Rich text is here",
  "classification": {
    "id": 12,
    "name": "138-cycles"
  },
  "properties": [{
    "id": 84436,
    "group": {
      "id": 19433,
      "name": "Suspension"
    },
    "name": "Fork turn",
    "description": "Fork turn defines bike suspension",
    "value": "129mm"
  }, {
    "id": 84436,
    "group": {
      "id": 19433,
      "name": "Suspension"
    },
    "name": "Fork turn",
    "description": "Fork turn defines bike suspension",
    "value": "129mm"
  }, {
    "id": 84436,
    "group": {
      "id": 19433,
      "name": "Suspension"
    },
    "name": "Fork turn",
    "description": "Fork turn defines bike suspension",
    "value": "129mm"
  }, {
    "id": 84436,
    "group": {
      "id": 19433,
      "name": "Suspension"
    },
    "name": "Fork turn",
    "description": "Fork turn defines bike suspension",
    "value": "129mm"
  }],
  "isGroup": true
}

I know how to build a case class for top level map:

case class ProductDocument(id: Long, name: String, annotation: String, description: String, isGroup: String) extends DocumentMap {
    ...
}

But I have no idea how to build values for vendor, category, properties, etc.

I want to define map for this JSON into one class file.

Answer 1

Each of the nested json objects should be defined as their own case classes, such that:

case class Vendor(id: Long, name: String)
case class ProductDocument(id: Long, ..., vendor: Vendor)

The properties would become a List of a Property case class:

case class Property(id: Long, group: PropertyGroup, name: String, description: String, value: String)
case class PropertyGroup(id: Long, name: String)
case class ProductDocument(id: Long, ..., properties: List[Property])

This assumes that you are using json4s serialization.

Answer 2

For scala-json mapping you have at least two options:

Salat - which originally provided ORM-like functionality for Mongo but also works for just JSON serialization (disclaimer: I work at Novus)

json4s - see section "Serialization"

The good news is that because the structure of Scala case classes are well-understood, you generally do not have to define "mapping" between your JSON fields and case class fields. You just write your case classes to match the structure / fields of your JSON document, as Arne Claassen has sketched out.