如何在不添加[JsonProperty]属性的情况下序列化JSON.NET中的静态属性

Question

是否可以使用JSON.NET序列化静态属性,而无需为每个属性添加[JsonProperty]属性.示例类:

public class Settings
    {
        public static int IntSetting { get; set; }
        public static string StrSetting { get; set; }

        static Settings()
        {
            IntSetting = 5;
            StrSetting = "Test str";
        }
    }

预期结果:

{
  "IntSetting": 5,
  "StrSetting": "Test str"
}

默认行为会跳过静态属性:

var x = JsonConvert.SerializeObject(new Settings(), Formatting.Indented);

Answer 1

您可以使用自定义合约解析程序执行此操作.具体来说,您需要子类化DefaultContractResolver并覆盖该GetSerializableMembers函数:

public class StaticPropertyContractResolver : DefaultContractResolver
{
    protected override List<MemberInfo> GetSerializableMembers(Type objectType)
    {
        var baseMembers = base.GetSerializableMembers(objectType);

        PropertyInfo[] staticMembers = 
            objectType.GetProperties(BindingFlags.Static | BindingFlags.Public);

        baseMembers.AddRange(staticMembers);

        return baseMembers;
    }
}

这里我们所做的就是调用基本实现GetSerializableMembers,然后将public static属性添加到我们的序列化成员列表中.

要使用它,您可以创建一个新JsonSerializerSettings对象并将其设置ContractResolver为以下实例StaticPropertyContractResolver:

var serializerSettings = new JsonSerializerSettings();

serializerSettings.ContractResolver = new StaticPropertyContractResolver();

现在,将这些设置传递给JsonConvert.SerializeObject所有应该工作:

string json = JsonConvert.SerializeObject(new Settings(), serializerSettings);

输出:

{
  "IntSetting": 5,
  "StrSetting": "Test str"
}

示例: https ://dotnetfiddle.net/pswTJW