Could not create an instance of type T. Type is an interface or abstract class_could not create an instance of type com.eriwen.gr

在开发中定义了一个抽象类，序列号抽象类的子类时，出现如下问题：

Could not create an instance of type TestJsonConvert.IDevice. Type is an interface or abstract class and cannot be instantiated.
1

解决方法：

在JsonSerializerSettings中设置TypeNameHandling 属性为TypeNameHandling.Auto即可。如下所示：

public abstract class Business
{
    public string Name { get; set; }
}

public class Hotel : Business
{
    public int Stars { get; set; }
}

public class Stockholder
{
    public string FullName { get; set; }
    public IList<Business> Businesses { get; set; }
}


Stockholder stockholder = new Stockholder
{
    FullName = "Steve Stockholder",
    Businesses = new List<Business>
    {
        new Hotel
        {
            Name = "Hudson Hotel",
            Stars = 4
        }
    }
};

string jsonTypeNameAll = JsonConvert.SerializeObject(stockholder, Formatting.Indented, new JsonSerializerSettings
{
    TypeNameHandling = TypeNameHandling.All
});

Console.WriteLine(jsonTypeNameAll);
// {
//   "$type": "Newtonsoft.Json.Samples.Stockholder, Newtonsoft.Json.Tests",
//   "FullName": "Steve Stockholder",
//   "Businesses": {
//     "$type": "System.Collections.Generic.List`1[[Newtonsoft.Json.Samples.Business, Newtonsoft.Json.Tests]], mscorlib",
//     "$values": [
//       {
//         "$type": "Newtonsoft.Json.Samples.Hotel, Newtonsoft.Json.Tests",
//         "Stars": 4,
//         "Name": "Hudson Hotel"
//       }
//     ]
//   }
// }

string jsonTypeNameAuto = JsonConvert.SerializeObject(stockholder, Formatting.Indented, new JsonSerializerSettings
{
    TypeNameHandling = TypeNameHandling.Auto
});

Console.WriteLine(jsonTypeNameAuto);
// {
//   "FullName": "Steve Stockholder",
//   "Businesses": [
//     {
//       "$type": "Newtonsoft.Json.Samples.Hotel, Newtonsoft.Json.Tests",
//       "Stars": 4,
//       "Name": "Hudson Hotel"
//     }
//   ]
// }

// for security TypeNameHandling is required when deserializing
Stockholder newStockholder = JsonConvert.DeserializeObject<Stockholder>(jsonTypeNameAuto, new JsonSerializerSettings
{
    TypeNameHandling = TypeNameHandling.Auto
});

Console.WriteLine(newStockholder.Businesses[0].GetType().Name);
// Hotel
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总结