容易解析JSON

简单解析JSON
我有下面一个字符串的JSON：我想获得那三个参数所对应的三个值，要怎么实现

{"type":"1061.jpg","EmployeeID":"c32df9b4-dd3b-46c5-8387-329b76356641","TaskID":"e52df9b4-ee3b-46c5-8387-329b76356641"}
[解决办法]
引用neNewtonsoft.Json.dll，上面的链接有。

 public class TT
    {
        public string Type { get; set; }
        public string EmployeeID { get; set; }
        public string TaskID { get; set; }
    }

static void Main(string[] args)
        {
            string json = "[{"type":"1061.jpg","EmployeeID":"c32df9b4-dd3b-46c5-8387-329b76356641","TaskID":"e52df9b4-ee3b-46c5-8387-329b76356641"}]";
            List<TT> list = JsonConvert.DeserializeObject<List<TT>>(json);

            foreach (TT info in list)
            {
                Console.WriteLine("type:" + info.Type + "\r\n EmployeeID:" + info.EmployeeID + "\r\n TaskID:" + info.TaskID);
            }
        }