//Model

public class Model
{
    public  datetime  DT{get;set;}
    public string Name{get;set;}
}
//---------
//xx.ashx文件

 List<Model> listmodel={xxxxxxx};

                        if (file.Count() != 0)
                        {
                            System.Web.Script.Serialization.JavaScriptSerializer json = new System.Web.Script.Serialization.JavaScriptSerializer();

                            string jsonStr = json.Serialize(listmodel);

                            context.Response.Write(jsonStr);
                        }

$.post("xxx.ashx", { "Name": cname, "Type": "customer" }, function(data) {

       if (data != "False") {
                 for(var i=0;i<data.length;i++)
                 {
                       var d=data[i].Time;  //这里获取的值，不再是时间。。据说要反序列化，额，求个函数       
                 }
             }
    },"JSON");
   

//获取序列化后的字符串，然后通过正则获取Date中的部分，在去进行转换。
    protected string JsonDateToDateString(string str)
    {
        Regex reg = new Regex(@"\\/Date\((\d+)\+\d+\)\\/");
        string time = reg.Match(str).Groups[1].Value;
        DateTime dt = new DateTime(1970, 1, 1);
        dt = dt.AddMilliseconds(long.Parse(time));
        dt = dt.ToLocalTime();
        return reg.Replace(str, dt.ToString("yyyy-MM-dd HH:mm:ss"));
    }

------解决方案--------------------

JScript code

取到的值是这样形式." /Date(1323100800000+0800)/"
 var Dtime =" /Date(1323100800000+0800)/";

                       var NewDtime = new Date(parseInt(Dtime.slice(6, 19)));

                       var Dyear = NewDtime.getFullYear();

                       var Dmonth = NewDtime.getMonth() + 1;

                       var Ddate = NewDtime.getDate();