EasyUI使用JSON保存数据

 1 <script type="text/javascript" src="easyui/jquery.min.js"></script> 2 <script type="text/Javascript" src="easyui/jquery.easyui.min.js"></script> 3 <script type="text/javascript" src="easyui/locale/easyui-lang-zh_CN.js"></script> 4 <script type="text/javascript" src="js/save.js"></script> 5 <link rel="stylesheet" type="text/CSS" href="easyui/themes/default/easyui.css" /> 6 <link rel="stylesheet" type="text/css" href="easyui/themes/icon.css" /> 7  8 　　<div> 9 　　　　<input type="text" id="txtFileNames" class="textbox" name="Vassets" style="width:110px;"/>10 　　　　<a href="#" class="easyui-linkbutton" iconCls="icon-search" onclick="objV.saves()">保存</a> 11 　　</div>

 1 objV = { 2         saves:function(){ 3             $.Ajax({ 4                 type: "POST", 5                 url: "Test.ashx", 6                 contentType: "application/json; charset=utf-8", 7                 data: JSON.stringify(GetJsonData()), 8                 dataType: "json", 9                 success: function (message) {10                     if (message > 0) {11                         alert("请求已提交！我们会尽快与您取得联系");12                     }13                 },14                 error: function (message) {15                     $("#request-PRocess-patent").html("提交数据失败！");16                 }17             });18         19         }20 21     };22 23 24     function GetJsonData() {25         var json = {26             "FTName": $("#txtFileNames").val()27         };28         return json;29     }

 1 　　　　　　  int num = 0; 2             context.Response.ContentType = "application/json"; 3             var data = context.Request; 4             var sr = new StreamReader(data.InputStream); 5             var stream = sr.ReadToEnd(); 6             var javaScriptSerializer = new JavaScriptSerializer(); 7             var PostedData = javaScriptSerializer.Deserialize<Model.FileType>(stream); 8             Dataaccess<Model.FileType> da = new DataAccess<Model.FileType>(); 9             10             11 12             try13             {14                 num = da.Add(PostedData, "FileType");15             }16             catch (Exception msg)17             {18                 context.Response.Write(msg.Message);19             }20 21             context.Response.ContentType = "text/plain";22             context.Response.Write(num);