2012-11-06 54 views
1

我有jqGrid和動態數據綁定的問題。 問題是,我總是得到Null參數頁面。我正在使用MVC3和jqGrid 4.4.1。jqGrid頁面返回null

var dataToSend = { 
      mrpId: selPRid, 
      mrpFilter: cookMainFilter, 
      divId: randomId, 
      chartId: chartId 
     }; 

     $("#gridMeasurementsTableView" + randomId).jqGrid({ 
      url: '/DataMonitor/CreateMeasurementTableView/', 
      datatype: "json", 
      mtype: 'POST', 
      postData: $.toDictionary(dataToSend), 
      jsonReader : { 
       root: "rows", 
       page: "page", 
       total: "total", 
       records: "records", 
       repeatitems: false, 
       cell: "cell", 
       id: "id" 
       }, 
      colNames: ['Date', 'Value'], 
      colModel: [ 
       { name: 'Timestamp', index: 'Timestamp', width: 280, 
        formatter: function (cellvalue, options, rowObject) { 
          return new Date(parseInt(cellvalue.substr(6, cellvalue.length - 8), 10)); 
         } 
       }, 
       { name: 'Value', index: 'Value', width: 200, align: 'center' } 
      ], 
      loadui: "disable", 
      loadtext: "Loading", 
      viewrecords: true, 
      height: "100%", 
      caption: "Measurements2", 
      pager: $("#gridpager" + randomId), 
      rowNum: 10, 
      sortname: 'Value', 
      sortorder: "asc", 
      rowList: [10, 50, 100] 
     }); 

和獲取數據的功能是這樣定義:

public ActionResult CreateMeasurementTableView(string sidx, string sord, int page, int rows, int mrpId, FilterMRPParameters mrpFilter, string divId, string chartId) 

BR, 揚

回答

0

嘗試改變postData: $.toDictionary(dataToSend)postData: dataToSend

+0

Tnx Oleg,現在工作。 – Jan

+0

@ user1802491:不客氣!如果問題解決了,您應該[「接受」](http://meta.stackexchange.com/a/5235/147495)答案。 – Oleg