使用catch all dictionary属性将json序列化为一个对象

Question

我想使用JSON.net反序列化为对象,但将未映射的属性放在字典属性中.可能吗？

比如给json,

 {one:1,two:2,three:3}

和c#类:

public class Mapped {
   public int One {get; set;}
   public int Two {get; set;}
   public Dictionary<string,object> TheRest {get; set;}
}

JSON.NET可以反序列化为值为1 = 1,2 = 1的实例,TheRest = Dictionary {{"three,3}}

Answer 1

The easiest way to do this is to use the JsonExtensionData attribute to define a catch all dictionary.

Example from the Json.Net documentation:

public class DirectoryAccount
{
    // normal deserialization
    public string DisplayName { get; set; }

    // these properties are set in OnDeserialized
    public string UserName { get; set; }
    public string Domain { get; set; }

    [JsonExtensionData]
    private IDictionary<string, JToken> _additionalData;

    [OnDeserialized]
    private void OnDeserialized(StreamingContext context)
    {
        // SAMAccountName is not deserialized to any property
        // and so it is added to the extension data dictionary
        string samAccountName = (string)_additionalData["SAMAccountName"];

        Domain = samAccountName.Split('\\')[0];
        UserName = samAccountName.Split('\\')[1];
    }

    public DirectoryAccount()
    {
        _additionalData = new Dictionary<string, JToken>();
    }
}

string json = @"{
  'DisplayName': 'John Smith',
  'SAMAccountName': 'contoso\\johns'
}";

DirectoryAccount account = JsonConvert.DeserializeObject<DirectoryAccount>(json);

Console.WriteLine(account.DisplayName);
// John Smith

Console.WriteLine(account.Domain);
// contoso

Console.WriteLine(account.UserName);
// johns

Answer 2

您可以创建一个CustomCreationConverter来执行您需要执行的操作。这是一个示例（相当丑陋，但演示了您可能想要如何实现这一点）：

namespace JsonConverterTest1
{
    public class Mapped
    {
        private Dictionary<string, object> _theRest = new Dictionary<string, object>();
        public int One { get; set; }
        public int Two { get; set; }
        public Dictionary<string, object> TheRest { get { return _theRest; } }
    }

    public class MappedConverter : CustomCreationConverter<Mapped>
    {
        public override Mapped Create(Type objectType)
        {
            return new Mapped();
        }

        public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
        {
            var mappedObj = new Mapped();
            var objProps = objectType.GetProperties().Select(p => p.Name.ToLower()).ToArray();

            //return base.ReadJson(reader, objectType, existingValue, serializer);
            while (reader.Read())
            {
                if (reader.TokenType == JsonToken.PropertyName)
                {
                    string readerValue = reader.Value.ToString().ToLower();
                    if (reader.Read())
                    {
                        if (objProps.Contains(readerValue))
                        {
                            PropertyInfo pi = mappedObj.GetType().GetProperty(readerValue, BindingFlags.IgnoreCase | BindingFlags.Public | BindingFlags.Instance);
                            var convertedValue = Convert.ChangeType(reader.Value, pi.PropertyType);
                            pi.SetValue(mappedObj, convertedValue, null);
                        }
                        else
                        {
                            mappedObj.TheRest.Add(readerValue, reader.Value);
                        }
                    }
                }
            }
            return mappedObj;
        }
    }

    public class Program
    {
        static void Main(string[] args)
        {
            string json = "{'one':1, 'two':2, 'three':3, 'four':4}";

            Mapped mappedObj = JsonConvert.DeserializeObject<Mapped>(json, new MappedConverter());

            Console.WriteLine(mappedObj.TheRest["three"].ToString());
            Console.WriteLine(mappedObj.TheRest["four"].ToString());
        }
    }
}

因此，反序列化 JSON 字符串后，mappedObj 的输出将是一个对象，其One和Two属性已填充，其他所有内容都放入Dictionary. 当然，我将“一”和“二”值硬编码为ints，但我认为这演示了您将如何处理此问题。

我希望这有帮助。

编辑：我更新了代码以使其更通用。我没有完全测试它，所以在某些情况下它可能会失败，但我认为它可以帮助您实现大部分目标。