C# JSONVERT反序列化抽象类数组
假设我有以下类结构(Building是一个抽象类): 然后,我创建了一个street对象,其中包含建筑物阵列中的几个平面:C# JSONVERT反序列化抽象类数组,c#,json.net,abstract-class,json-deserialization,C#,Json.net,Abstract Class,Json Deserialization,假设我有以下类结构(Building是一个抽象类): 然后,我创建了一个street对象,其中包含建筑物阵列中的几个平面: Flat f1 = new Flat { Name = "Flat 1", WhichFloor = 1 }; Flat f2 = new Flat { Name = "Flat 2", WhichFloor = 2 }; Street street = new Street { StreetName = "Street Name", Buildings
Flat f1 = new Flat { Name = "Flat 1", WhichFloor = 1 };
Flat f2 = new Flat { Name = "Flat 2", WhichFloor = 2 };
Street street = new Street
{
StreetName = "Street Name",
Buildings = new[] { f1, f2 }
};
然后使用JsonConvert序列化对象:
var toJson = JsonConvert.SerializeObject(street);
现在我想将json转换回street对象:
var fromJson = JsonConvert.DeserializeObject<Street>(toJson);
var fromJson=JsonConvert.DeserializeObject(toJson);
此操作失败,出现以下错误:
“无法创建Building类型的实例。type是接口或抽象类,无法实例化。路径'Buildings[0]。WhichFloor'”
如何告诉JsonConvert类建筑应该是一组平面?根据@Evk的共享链接,在序列化和反序列化时,您应该尝试将
TypeNameHandling
设置为TypeNameHandling.Auto
:
var toJson = JsonConvert.SerializeObject(street, Newtonsoft.Json.Formatting.Indented, new JsonSerializerSettings
{
TypeNameHandling = TypeNameHandling.Auto
});
var fromJson = JsonConvert.DeserializeObject<Street>(toJson, new JsonSerializerSettings
{
TypeNameHandling = TypeNameHandling.Auto
});
var-toJson=JsonConvert.SerializeObject(street,Newtonsoft.Json.Formatting.Indented,new-JsonSerializerSettings
{
TypeNameHandling=TypeNameHandling.Auto
});
var fromJson=JsonConvert.DeserializeObject(toJson,新的JsonSerializerSettings
{
TypeNameHandling=TypeNameHandling.Auto
});
公共抽象类JsonCreationConverter:JsonConverter
{
受保护的抽象T Create(类型objectType,JObject-JObject);
公共覆盖布尔CanConvert(类型objectType)
{
返回类型(T)==objectType;
}
公共重写对象ReadJson(JsonReader reader,类型objectType,
对象存在值,JsonSerializer序列化程序)
{
尝试
{
var jObject=jObject.Load(读卡器);
var target=Create(objectType,jObject);
填充(jObject.CreateReader(),目标);
回报目标;
}
捕获(JsonReaderException)
{
返回null;
}
}
公共重写void WriteJson(JsonWriter writer,对象值,
JsonSerializer(序列化程序)
{
抛出新的NotImplementedException();
}
}
现在实现这个接口
public class SportActivityConverter : JsonCreationConverter<BaseSportActivity>
{
protected override BaseSportActivity Create(Type objectType, JObject jObject)
{
BaseSportActivity result = null;
try
{
switch ((ESportActivityType)jObject["activityType"].Value<int>())
{
case ESportActivityType.Football:
result = jObject.ToObject<FootballActivity>();
break;
case ESportActivityType.Basketball:
result = jObject.ToObject<BasketballActivity>();
break;
}
}
catch(Exception ex)
{
Debug.WriteLine(ex);
}
return result;
}
}
公共类SportActivityConverter:JsonCreationConverter
{
受保护的重写BaseSportActivity创建(类型objectType,JObject JObject)
{
BaseSportActivity结果=空;
尝试
{
开关((ESportActivityType)作业对象[“activityType”].Value())
{
案例ESportActivityType.足球:
结果=jObject.ToObject();
打破
案例ESportActivityType.篮球:
结果=jObject.ToObject();
打破
}
}
捕获(例外情况除外)
{
Debug.WriteLine(ex);
}
返回结果;
}
}
你需要这个:我发现最好的结果来自于typenameholling。所有这些
public abstract class JsonCreationConverter<T> : JsonConverter
{
protected abstract T Create(Type objectType, JObject jObject);
public override bool CanConvert(Type objectType)
{
return typeof(T) == objectType;
}
public override object ReadJson(JsonReader reader,Type objectType,
object existingValue, JsonSerializer serializer)
{
try
{
var jObject = JObject.Load(reader);
var target = Create(objectType, jObject);
serializer.Populate(jObject.CreateReader(), target);
return target;
}
catch (JsonReaderException)
{
return null;
}
}
public override void WriteJson(JsonWriter writer, object value,
JsonSerializer serializer)
{
throw new NotImplementedException();
}
}
public class SportActivityConverter : JsonCreationConverter<BaseSportActivity>
{
protected override BaseSportActivity Create(Type objectType, JObject jObject)
{
BaseSportActivity result = null;
try
{
switch ((ESportActivityType)jObject["activityType"].Value<int>())
{
case ESportActivityType.Football:
result = jObject.ToObject<FootballActivity>();
break;
case ESportActivityType.Basketball:
result = jObject.ToObject<BasketballActivity>();
break;
}
}
catch(Exception ex)
{
Debug.WriteLine(ex);
}
return result;
}
}