Json数据时间格式的处理

方法:
 1 using Newtonsoft.Json;
 2 using Newtonsoft.Json.Converters;//需引入Newtonsoft.Json.dll
 3 public class ConvertHelper
 4     {
 5         /// <summary>
 6         /// 将对象转成json格式并格式化日期:yyyy-MM-dd
 7         /// </summary>
 8         /// <param name="jsonObject"></param>
 9         /// <param name="dateType"></param>
10         /// <returns></returns>
11         public static string BuildJsonDateString(object jsonObject, string dateType = "yyyy-MM-dd")
12         {
13             var timeConverter = new IsoDateTimeConverter { DateTimeFormat = dateType };
14             return JsonConvert.SerializeObject(jsonObject, Newtonsoft.Json.Formatting.Indented, timeConverter);
15         }
16 
17         /// <summary>
18         /// 将对象转成json格式并格式化时间:yyyy-MM-dd HH:mm:ss
19         /// </summary>
20         /// <param name="jsonObject"></param>
21         /// <param name="dateType"></param>
22         /// <returns></returns>
23         public static string BuildJsonDateTimeString(object jsonObject, string dateType = "yyyy-MM-dd HH:mm:ss")
24         {
25             var timeConverter = new IsoDateTimeConverter { DateTimeFormat = dateType };
26             return JsonConvert.SerializeObject(jsonObject, Newtonsoft.Json.Formatting.Indented, timeConverter);
27         }
28     }
View Code

 

调用:

1 public string GetList()
2         {
3             News_BLL bll = new News_BLL();
4             DataTable dt = bll.GetList("").Tables[0];
5             var list = Common.ConvertHelper<Model.News_Model.News_ITM>.ConvertToList(dt);
6             return Common.ConvertHelper.BuildJsonDateTimeString(list);
7         }
View Code

 

posted @ 2015-11-17 17:29  熊大大-  阅读(525)  评论(0编辑  收藏  举报