使用GSON将JSON对象转换为具有不同格式的Java对象

Question

I have a response that returns a json object in following format:

{
    "playerId": "001",
    "name": "michel",
    "age": 21,
    "nation": "USA",
    "ratings": [
        {
            "type": "speed",
            "score": "0121"
        },
        {
            "type": "accuracy",
            "score": "85"
        }
    ],
    "teaminfo": {
        "teamName": "HON",
        "isValid": "true"
    }
}

and I have a Java Class as :

public class MyRider {
    public String playerId;
    public String name;
    public int age;
    public String speed;
    public String accuracy;
    public String teamName;
    public String isValid;
    //getter, setter... 
}

I want to map the JSON object into Java object using GSON.

I tried using JsonDeserializationContext deserialize, and it returned null for the nested values in JSON.

Answer 1

Without custom deserializer

If you cannot change the JSON to return exactly what you want, I suggest you create classes to match it:

MyRider:

public class MyRider {
    private String playerId;
    private String name;
    private int age;
    private String nation;
    private List<Rating> ratings;
    private TeamInfo teaminfo;

    // getters, setters, toString override
}

Rating:

public class Rating {
    private String type;
    private String score;

    // getters, setters, toString override
}

TeamInfo:

private static class TeamInfo {
    private String teamName;
    private String isValid;

    // getters, setters, toString override
}

Then simply deserialize as normal:

MyRider rider = gson.fromJson(json, MyRider.class);

If you need exactly the fields you've specified in MyRider in your question, consider a transformer class to map the full class above to your needs.

With custom deserializer

It's also possible to do this with a custom deserializer, but slightly pointless as GSON provides the normal mapping for you which you can then adapt.

Here is an example with a deserializer:

public class MyRiderDeserializer implements JsonDeserializer<MyRider> {

    @Override
    public MyRider deserialize(JsonElement json, Type typeOfT, 
                               JsonDeserializationContext context)
                               throws JsonParseException {
        MyRider rider = new MyRider();
        if(json.isJsonObject()) {
            JsonObject riderObj = json.getAsJsonObject();

            rider.setPlayerId(riderObj.get("playerId").getAsString());
            rider.setName(riderObj.get("name").getAsString());
            rider.setAge(riderObj.get("age").getAsInt());

            JsonArray ratingsArray = riderObj.get("ratings").getAsJsonArray();
            for(JsonElement ratingElem : ratingsArray) {
                JsonObject ratingObj = ratingElem.getAsJsonObject();
                String type = ratingObj.get("type").getAsString();
                switch(type) {
                    case "speed":
                        rider.setSpeed(ratingObj.get("score").getAsString());
                        break;
                    case "accuracy":
                        rider.setAccuracy(ratingObj.get("score").getAsString());
                        break;
                    default:
                        break;
                }
            }

            JsonObject teamInfo = riderObj.get("teaminfo").getAsJsonObject();
            rider.setTeamName(teamInfo.get("teamName").getAsString());
            rider.setIsValid(teamInfo.get("isValid").getAsString());

        }
        return rider;
    }
}

Note this does not include any checks to validate whether the properties are actually there and is the simplest possible custom deserializer I could think of. To use it, you must register the type adapter at Gson creation time:

Gson gson = new GsonBuilder()
                .registerTypeAdapter(MyRider.class, new MyRiderDeserializer())
                .create();

MyRider myRider = gson.fromJson(reader, MyRider.class);