Deserialize a JSON payload to object base on JSON integer property

Question

I have below classes:

public class Result<T> {
    public int code;
    public Object meta;
    public T data;
}

public class User {
    public int id;
    public String name;
}

public class Error {
    public String field;
    public String message;
}

I want to deserialize a JSON payload based on code field. If code >= 10, return Result<ArrayList<Error>>, otherwise return Result<User>

Currently, I map JSON to Result<Object> first, then check the code field. Based on that value I make second map to desired object.

ObjectMapper mapper = new ObjectMapper();
Result<Object> tempResult = mapper.readValue(json, new TypeReference<Result<Object>>() {});

if (tempResult.code < 10) {    
    Result<User> result = mapper.readValue(json, new TypeReference<Result<User>>() {});
    return result;
} else {
    Result<ArrayList<Error>> result = mapper.readValue(json, new TypeReference<Result<ArrayList<Error>>>() {});
    return result;
}

Is there an elegant way to do this without deserializing it 2 times?

Michał Ziober · Accepted Answer · 2020-09-17 10:07:43Z

You need to implement custom TypeIdResolver:

class UserTypeIdResolverBase extends TypeIdResolverBase {

    @Override
    public String idFromValue(Object value) {
        throw new IllegalStateException("Not implemented!");
    }

    @Override
    public String idFromValueAndType(Object value, Class<?> suggestedType) {
        throw new IllegalStateException("Not implemented!");
    }

    @Override
    public JsonTypeInfo.Id getMechanism() {
        return JsonTypeInfo.Id.CUSTOM;
    }

    @Override
    public JavaType typeFromId(DatabindContext context, String id) {
        if (Integer.parseInt(id) < 10) {
            return context.getTypeFactory().constructType(new TypeReference<Result<User>>() {});
        }
        return context.getTypeFactory().constructType(new TypeReference<Result<List<Error>>>() {});
    }
}

and declare it for a Result class:

@JsonTypeInfo(property = "code", use = JsonTypeInfo.Id.CUSTOM, visible = true)
@JsonTypeIdResolver(UserTypeIdResolverBase.class)
class Result<T>

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