8 окт. 2019 г. · 2 Answers 2 · Paste your JSON data to left column. · Select the Source Type -> JSON · Select the Serialization framework -> Just Types · Choose ...

24 мая 2022 г. · I have the following JSON structure that I getting from the server: { "a": { "something": "something", "something2": { "test": "12345", ...

30 янв. 2017 г. · I'm receiving a quite deep JSON object string from a service which I must parse to a JSON object and then map it to classes. How can I transform ...

11 янв. 2016 г. · ... Kotlin objects when you do no have a default constructor. Your first sample of the code: val dtos = mapper.readValue(json, List::class.java).

22 мая 2017 г. · Details · Gson version: 2.8.5 · Android Studio 3.1.4 · Kotlin version: 1.2.60. Solution. Create any class data and inherit ...

25 сент. 2019 г. · In 2019 no-one is really parsing JSON manually. It's much easier to use Gson library. It takes as an input your object and spits out JSON ...

19 янв. 2024 г. · Android How to convert this JSON into object of data class with Moshi Kotlin? · This is your json file or are you get it from somewhere? You can ...

2 июл. 2017 г. · This can be done with Klaxon. With this you can easily read the Json data as JsonObject which is actually a MutableMap . val json: JsonObject = ...

30 янв. 2022 г. · With kotlinx.serialization you can parse JSON String into a JsonElement : val json: Map<String, JsonElement> = Json.

5 янв. 2023 г. · I am trying to deserialize nested JSON into a single DTO in a Spring Boot application. The JSON: { "values":[ { "name":"test", "map":{ "d1 ...