为JSON提要定义嵌套类的最佳方法

时间:2012-08-13 14:33:25

标签: c# json class json.net

我目前正在使用c#中的Facebook API,使用NewtonSoft JSON库来使用所有返回的API数据。

返回用户页面列表我发现自己在返回的JSON中为属性创建了一个关闭类,以便对其进行序列化。

现在我有这个:

    public class FacebookPage
    {
        public string id { get; set; }
        public string name { get; set; }
        public string link { get; set; }
        public string category { get; set; }
        public bool is_published { get; set; }
        public bool can_post { get; set; }
        public int likes { get; set; }
        public FacebookPageLocation location { get; set; }
        public string phone { get; set; }
        public int checkins { get; set; }
        public string picture { get; set; }
        public FacebookPageCover cover { get; set; }
        public string website { get; set; }
        public int talking_about_count { get; set; }
        public string access_token { get; set; }
    }
    public class FacebookPageLocation
    {
        public decimal latitude { get; set; }
        public decimal longitude { get; set; }
    }
    public class FacebookPageCover
    {
        public string cover_id { get; set; }
        public string source { get; set; }
        public int offset_y { get; set; }
    }

似乎必须有更好的方法来做到这一点。我可以用Dictionary替换FacebookPageLocation,但是如何更换FacebookCoverPage呢?

理想情况下,我希望能够以一种漂亮的嵌套格式声明它,就像这样

    public class FacebookPage
    {
        id = string,
        name = string.
        link = string,
        category = string,
        is_published = bool,
        can_post = bool,
        likes = int,
        location = 
        {
            latitude = decimal,
            longtitude = decimal
        },
        phone = string,
        checkins = int,
        picture = string,
        cover  = 
        {
            cover_id = string,
            source = string,
            offset_y = int
        }
        website = string,
        talking_about_count = int,
        access_token = string
    }

我意识到这在实践中不起作用,但是有什么东西可以宣布这些类更整洁吗?还是不必要的?

1 个答案:

答案 0 :(得分:0)

我会使用dynamic而不是声明很多类。对于下面的样本json

{
  "name": "joe",
  "id": 1151,
  "location": {
    "lat": 180.0,
    "lng": -180.0
  }
}

-

代码将是

dynamic obj = JsonConvert.DeserializeObject(json);
Console.WriteLine("{0} {1} {2}", obj.id, obj.name, obj.location.lat);