我希望实施https://github.com/HeadspringLabs/Enumeration。目前,当我尝试序列化/反序列化Enumeration时,它将它序列化为一个复杂的对象。例如,对象颜色:
public class Color : Enumeration<Color, int>
{
public static readonly Color Red = new Color(1, "Red");
public static readonly Color Blue = new Color(2, "Blue");
public static readonly Color Green = new Color(3, "Green");
private Color(int value, string displayName) : base(value, displayName) { }
}
将序列化为
{
"_value": 2,
"_displayName": "Something"
}
在像这样的复杂对象中:
public class OtherClass
{
public Color Color {get; set;}
public string Description {get; set;}
}
它会像这样序列化:
{
"Description" : "Other Description",
"Color" :
{
"_value": 2,
"_displayName": "Something"
}
}
有没有办法让json转换序列化这样的复杂对象:
{
"Description" : "Other Description",
"Color" : 2
}
我可以使用Enumeration类中的FromValue方法从值中创建正确的Color对象。我似乎无法使json转换将属性值作为Color对象的“值”。
我可以用什么方式编写转换器的WriteJson和Create方法来实现?
public class EnumerationConverter : JsonCreationConverter<IEnumeration>
{
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
}
protected override IEnumeration Create(Type objectType, JObject jObject)
{
}
}
答案 0 :(得分:2)
您可以为Headspring.Enumeration<T, int>
- 派生类创建一个通用转换器,如下所示:
class EnumerationConverter<T> : JsonConverter where T : Headspring.Enumeration<T, int>
{
public override bool CanConvert(Type objectType)
{
return objectType == typeof(T);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
int val = Convert.ToInt32(reader.Value);
return Headspring.Enumeration<T, int>.FromValue(val);
}
public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
var enumVal = (Headspring.Enumeration<T, int>)value;
writer.WriteValue(enumVal.Value);
}
}
要使用转换器,请在您的枚举类中添加[JsonConverter]
属性,如下所示:
[JsonConverter(typeof(EnumerationConverter<Color>))]
public class Color : Headspring.Enumeration<Color, int>
{
public static readonly Color Red = new Color(1, "Red");
public static readonly Color Blue = new Color(2, "Blue");
public static readonly Color Green = new Color(3, "Green");
private Color(int value, string displayName) : base(value, displayName) { }
}
这是一个往返演示:https://dotnetfiddle.net/CZsQab