Decoding with Gson: Say No to Wrapper Classes

Working on deserializing the JSON data provided using Gson in Kotlin.

{
  "things": [
    {
      "name": "Thing1"
    },
    {
      "name": "Thing2"
    }
  }
}

Currently, there are two classes involved in deserialization;

  1. ThingWrapper - Contains a property called 'Things' which is a List
  2. Thing - Contains a name property

The service returns the List<Thing> from 'things' without exposing the ThingWrapper.

fun getThings(): List<Thing> {
   return Gson().fromJson(json, ThingWrapper::class.java).things
}

Is it possible to deserialize the data using Gson without the need for a ThingWrapper class at all?

Thanks, John

Answer №1

If you're aiming to remove the Wrapper class, one alternative is to utilize the JSONObject class for parsing the object and then extracting the JSON array from it.

val items: List<Item> = Gson().fromJson(
            JSONObject(json).getJSONArray("items").toString(),
            object : TypeToken<List<Item>>() {}.type)
Log.i("result", items.toString())
<

I/result: [Item(name=Item1), Item(name=Item2)]

This approach proved successful in my case.

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