JSON.net：如何在不使用默认构造函数的情况下反序列化

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)
{
    ...
}

构造函数参数名必须与JSON对象的相应属性名匹配(忽略大小写)，这一点很重要，这样才能正常工作.然而，对于对象的每个属性，您不一定都必须有一个构造函数参数.对于构造函数参数未涵盖的JSON对象属性，JSON.Net将try 使用公共属性访问器(或标记为[JsonProperty]的属性/字段)在构建对象后填充该对象.

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);