Method:
1 usingNewtonsoft.json;2 usingNewtonsoft.Json.Converters;//need to introduce Newtonsoft.Json.dll3 Public classConverthelper4 {5 /// <summary>6 ///Convert an object to JSON format and format date: Yyyy-mm-dd7 /// </summary>8 /// <param name= "Jsonobject" ></param>9 /// <param name= "Datetype" ></param>Ten /// <returns></returns> One Public Static stringBuildjsondatestring (ObjectJsonobject,stringDatetype ="YYYY-MM-DD") A { - varTimeConverter =Newisodatetimeconverter {DateTimeFormat =Datetype}; - returnJsonconvert.serializeobject (Jsonobject, Newtonsoft.Json.Formatting.Indented, timeconverter); the } - - /// <summary> - ///Convert object to JSON format and format time: Yyyy-mm-dd HH:mm:ss + /// </summary> - /// <param name= "Jsonobject" ></param> + /// <param name= "Datetype" ></param> A /// <returns></returns> at Public Static stringBuildjsondatetimestring (ObjectJsonobject,stringDatetype ="YYYY-MM-DD HH:mm:ss") - { - varTimeConverter =Newisodatetimeconverter {DateTimeFormat =Datetype}; - returnJsonconvert.serializeobject (Jsonobject, Newtonsoft.Json.Formatting.Indented, timeconverter); - } -}
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Call:
1 Public stringGetList ()2 {3NEWS_BLL BLL =NewNEWS_BLL ();4DataTable dt = BLL. GetList (""). tables[0];5 varList = Common.converthelper<model.news_model.news_itm>. Converttolist (DT);6 returnCommon.ConvertHelper.BuildJsonDateTimeString (list);7}
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Processing of JSON data time format