1

我想要我的类类型的 JSON 表示，以便我可以将它们插入客户端模式。这就是我的服务器端 C# 类的样子：

public class Person
{
    public int ID { get; set; }

    [Required(ErrorMessage = "Name is required.")]
    public string Name { get; set; }

    public DateTime Birthday { get; set; }

    public bool IsMarried { get; set; }
}

我如何将它序列化为这个？：

{
    fields: {
        "ID": {
            type: "number"
        },
        "Name": {
            type: "string",
            validation: {
                required: true,
                message: "Name is required.",
            }
        },
        "Birthday": {
            type: "date"
        },
        "IsMarried": {
            type: "bool"
        }
    }
}
4

2 回答 2

2

JSON.NET 库中没有任何现成的（AFAIK），但使用一些反射代码实现它相当简单：

public class StackOverflow_19336832
{
    public class RequiredAttribute : Attribute
    {
        public string ErrorMessage { get; set; }
    }

    public class Person
    {
        public int ID { get; set; }

        [Required(ErrorMessage = "Name is required.")]
        public string Name { get; set; }

        public DateTime Birthday { get; set; }

        public bool IsMarried { get; set; }
    }

    static JObject GetSchema(Type type)
    {
        var result = new JObject();
        foreach (var prop in type.GetProperties())
        {
            var name = prop.Name;
            var propType = prop.PropertyType;
            var field = new JObject();
            result.Add(name, field);
            switch (Type.GetTypeCode(propType))
            {
                case TypeCode.Boolean:
                    field.Add("type", "bool");
                    break;
                case TypeCode.Int16:
                case TypeCode.Int32:
                case TypeCode.Int64:
                case TypeCode.UInt16:
                case TypeCode.UInt32:
                case TypeCode.UInt64:
                case TypeCode.Double:
                case TypeCode.Single:
                    field.Add("type", "number");
                    break;
                case TypeCode.String:
                    field.Add("type", "string");
                    break;
                case TypeCode.DateTime:
                    field.Add("type", "date");
                    break;
                default:
                    throw new ArgumentException("Don't know how to generate schema for property with type " + propType.FullName);
            }

            var req = prop.GetCustomAttributes(typeof(RequiredAttribute), false).FirstOrDefault() as RequiredAttribute;
            if (req != null)
            {
                var validation = new JObject();
                field.Add("validation", validation);
                validation.Add("required", true);
                if (!string.IsNullOrEmpty(req.ErrorMessage))
                {
                    validation.Add("message", req.ErrorMessage);
                }
            }
        }

        return result;
    }

    public static void Test()
    {
        var schema = GetSchema(typeof(Person));
        Console.WriteLine(schema);
    }
}
于 2013-10-12T18:27:51.873 回答
-1

在这里查看一些关于如何使用JavaScriptSerializer来完成工作的想法。还有其他序列化程序，但这个运行良好，链接提供了如何实现和使用它的示例。

http://msdn.microsoft.com/en-us/library/system.web.script.serialization.javascriptserializer.aspx

于 2013-10-12T17:09:00.880 回答