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在C#中覆盖Json属性名称

更新时间:2023-02-17 21:26:25

您将必须覆盖DefaultContractResolver并实现您自己的机制来提供PropertyName(以JSON).我将提供完整的示例代码,以显示运行时生成的PropertyName的反序列化和序列化.当前,它会将Test字段修改为Test5(在所有型号中).您应该实现自己的机制(使用属性,保留名称,表等).

You'll have to override DefaultContractResolver and implement your own mechanism to provide the PropertyName (in JSON). I will provide a full example code to show deserialization and serialization with a runtime generated PropertyName. Currently, it modifies the Test field to Test5 (in all models). You should implement your own mechanism (using an attribute, a reserved name, a table or whatever.

class Program
{
    static void Main(string[] args)
    {
        var customer = new Customer() {Email = "asd@asd.com", Test = "asdasd"};
        var a = Serialize(customer, false);
        var b = Serialize(customer, true);
        Console.WriteLine(a);
        Console.WriteLine(b);

        var desA = Deserialize<Customer>(a, false);
        var desB = Deserialize<Customer>(b, true);

        Console.WriteLine("TestA: {0}", desA.Test);
        Console.WriteLine("TestB: {0}", desB.Test);

    }

    static string Serialize(object obj, bool newNames)
    {
        JsonSerializerSettings settings = new JsonSerializerSettings();
        settings.Formatting = Formatting.Indented;
        if (newNames)
        {
            settings.ContractResolver = new CustomNamesContractResolver();
        }

        return JsonConvert.SerializeObject(obj, settings);
    }
    static T Deserialize<T>(string text, bool newNames)
    {
        JsonSerializerSettings settings = new JsonSerializerSettings();
        settings.Formatting = Formatting.Indented;
        if (newNames)
        {
            settings.ContractResolver = new CustomNamesContractResolver();
        }

        return JsonConvert.DeserializeObject<T>(text, settings);
    }
}
class CustomNamesContractResolver : DefaultContractResolver
{
    protected override IList<JsonProperty> CreateProperties(System.Type type, MemberSerialization memberSerialization)
    {
        // Let the base class create all the JsonProperties 
        // using the short names
        IList<JsonProperty> list = base.CreateProperties(type, memberSerialization);

        // Now inspect each property and replace the 
        // short name with the real property name
        foreach (JsonProperty prop in list)
        {
            if (prop.UnderlyingName == "Test") //change this to your implementation!
                prop.PropertyName = "Test" + 5;

        }

        return list;
    }
}

public class Customer
{
    [JsonProperty(PropertyName = "email")]
    public string Email { get; set; }

    public string Test { get; set; }

}

输出:

{
  "email": "asd@asd.com",
  "Test": "asdasd"
}
{
  "email": "asd@asd.com",
  "Test5": "asdasd"
}
TestA: asdasd
TestB: asdasd

如您所见,

如您所见,当我们使用Serialize(..., false)时,该字段的名称为Test;当我们使用Serialize(..., true)时,该字段的名称为Test5.这也适用于反序列化.

As you see, when we use Serialize(..., false) - the field's name is Test and when we use Serialize(..., true) - the field's name is Test5, as expected. This also works for deserialization.

我已将此答案用作答案: https://***.com/a/20639697/773879

I have used this answer as insperation for my answer: https://***.com/a/20639697/773879