如何提供超鏈接到JQgrid中的列數據並根據點擊數據導航到相應的Url？

Question

我有一個jqGrid的包含有僅2 columns..I我貼我的代碼，我在下面試圖..

var UserArr = new Array(); 
function Grid() 
{ 
$("#users_grid").jqGrid({ 

      colNames: ['Site_Name', 'Details'], 
      colModel: [{ name: 'Site_Name', index: 'Site_Name', width: 130,editable: false, sortable: false,formatter: 'showlink', formatoptions: { baseLinkUrl: 'url of respective site i have clicked'}}, 
      { name: 'Details', index: 'Details', width: 400, editable: false, sortable: false }], 

      width: 400, 
      height: 'auto', 
      multiselect: true 

     }); 

     var postJSONData = JSON.stringify({ 'parentitem': parent,'childitem':child }); 

     $.ajax({ 
      type: 'POST', 
      data: postJSONData, 
      url: 'ManageAssetService.asmx/DisplayGridData', 
      dataType: 'json', 
      async: false, 
      contentType: 'application/json; charset=utf-8', 
      success: function success(response) { 

       UserArr = response.d; 

      }, 
      error: function failure(response) { 
      alert(response.message); 
       alert('failed to fetch user details'); 
      } 
     }); 

     var mydata; 

     for (var i = 0; i <5; i++) { 

      mydata = {}; 

      mydata.Url= UserArr[i].Url; 
      mydata.Details= UserArr[i].Details; 

      $("#users_grid").jqGrid('addRowData', 'GridData_Row_' + (i + 1), mydata); 


     } 
} 

我就會顯示該網站名稱以及有關在jqGrid的一些細節。現在Site_Name將是一個超鏈接，當我點擊它應該重定向到相應的網址..我怎麼能實現該..而且我正在動態添加網格數據..我應該在哪裏給Site_Name列數據各自的URL和我怎麼可能有鏈接呢..

請幫助..

Answer 1

我有我的SOLUT離子這樣的..

的柱的jqGrid shold這樣被定義：

colNames: ['Site_Name', 'Details'], 
      colModel: [{ name: 'Site_Name', index: 'Site_Name', width: 130,editable: false, sortable: false,formatter: 'showlink', formatoptions: { baseLinkUrl: 'javascript:', showAction: "Link('", addParam: "');"} }, 
      { name: 'Details', index: 'Details', width: 400, editable: false, sortable: false }], 

JavaScript函數：

function Link(id) { 

    var row = id.split("="); 
    var row_ID = row[1]; 
    var sitename= $("#users_grid").getCell(row_ID, 'Site_Name'); 
    var url = "http://"+sitename; // sitename will be like google.com or yahoo.com 

    window.open(url); 


} 

那它..