How to deserialize only some fields in JSON?

Question

I'm using Gson to extraxt some fields. By the way I don't want to create a class due to the fact that I need only one value in all JSON response. Here's my response:

{
    "result": {
        "name1": "value1",
        "name2": "value2",
    },
    "wantedName": "wantedValue"
}

I need wantedValue but I don't want to create the entire class for deserialization. Is it possible to achieve this using Gson?

Answer 1

If you need one field only, use JSONObject.

import org.json.JSONException;
import org.json.JSONObject;


public class Main { 
public static void main(String[] args) throws JSONException  {

    String str = "{" + 
            "    \"result\": {" + 
            "        \"name1\": \"value1\"," + 
            "        \"name2\": \"value2\"," + 
            "    }," + 
            "    \"wantedName\": \"wantedValue\"" + 
            "}";

    JSONObject jsonObject = new JSONObject(str);

    System.out.println(jsonObject.getString("wantedName"));
}

Output:

wantedValue