c# - JSON.net: how to deserialize without using the default constructor?

Question

I have a class that has a default constructor and also an overloaded constructor that takes in a set of parameters. These parameters match to fields on the object and are assigned on construction. At this point i need the default constructor for other purposes so i would like to keep it if i can.

My Problem: If I remove the default constructor and pass in the JSON string, the object deserializes correctly and passes in the constructor parameters without any issues. I end up getting back the object populated the way I would expect. However, as soon as I add the default constructor into the object, when i call JsonConvert.DeserializeObject<Result>(jsontext) the properties are no longer populated.

At this point I have tried adding new JsonSerializerSettings(){CheckAdditionalContent = true} to the deserialization call. That did not do anything.

Another note: the constructor parameters do match the names of the fields exactly except that the parameters are start with a lowercase letter. I wouldn't think this would matter since, like i mentioned, the deserialization works fine with no default constructor.

Here is a sample of my constructors:

public Result() { }

public Result(int? code, string format, Dictionary<string, string> details = null)
{
    Code = code ?? ERROR_CODE;
    Format = format;

    if (details == null)
        Details = new Dictionary<string, string>();
    else
        Details = details;
}

Answer 1

Json.Net prefers to use the default (parameterless) constructor on an object if there is one. If there are multiple constructors and you want Json.Net to use a non-default one, then you can add the [JsonConstructor] attribute to the constructor that you want Json.Net to call.

[JsonConstructor]
public Result(int? code, string format, Dictionary<string, string> details = null)
{
    ...
}

It is important that the constructor parameter names match the corresponding property names of the JSON object (ignoring case) for this to work correctly. You do not necessarily have to have a constructor parameter for every property of the object, however. For those JSON object properties that are not covered by the constructor parameters, Json.Net will try to use the public property accessors (or properties/fields marked with [JsonProperty]) to populate the object after constructing it.

If you do not want to add attributes to your class or don't otherwise control the source code for the class you are trying to deserialize, then another alternative is to create a custom JsonConverter to instantiate and populate your object. For example:

class ResultConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return (objectType == typeof(Result));
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        // Load the JSON for the Result into a JObject
        JObject jo = JObject.Load(reader);

        // Read the properties which will be used as constructor parameters
        int? code = (int?)jo["Code"];
        string format = (string)jo["Format"];

        // Construct the Result object using the non-default constructor
        Result result = new Result(code, format);

        // (If anything else needs to be populated on the result object, do that here)

        // Return the result
        return result;
    }

    public override bool CanWrite
    {
        get { return false; }
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

Then, add the converter to your serializer settings, and use the settings when you deserialize:

JsonSerializerSettings settings = new JsonSerializerSettings();
settings.Converters.Add(new ResultConverter());
Result result = JsonConvert.DeserializeObject<Result>(jsontext, settings);