2015-07-28 62 views
0

我已經創建了一個json對象,並且想要讀取內容並將值放入dataTable中。但我無法讀取Json對象的內容。無法從Json對象讀取字段的值

下面是我的javascript函數:

finalGrid : function(data){ 
      console.log("final Grid"); 
      var strJson = JSON.stringify(data); 
      var dataJson = JSON.parse(strJson); 
      console.log(dataJson); 
      var table = $('#' + Dcn.tableId).dataTable(); 
      var tableObj = table.api(); 
      var rows = []; 
      var field = Dcn.tableField; 
      var count=0; 
      $.each(dataJson, function(index, obj) { 
       var row = []; 
       row[field.dcnRefNo.index] = obj[field.dcnRefNo.name]; 
       row[field.name.index] = obj[field.name.name]"; 
       row[field.package.index] = obj[field.package.name]; 
       row[field.priority.index] = obj[field.priority.name]; 

       row.push("<button class='btn btn-labeled btn-success project-edit' " + 
         "onclick=\"Dcn.List.bindDcnStatusUpdate('"+obj["_id"]+"')\"> " + 
         "<span class='btn-label'><i class='fa fa-edit'></i></span>View Dcn</button>"); 
      rows.push(row); 
      count = count+1; 
      console.log(count); 
     }); 

     tableObj.rows.add(rows).draw(); 
     $('#' + Dcn.tableId).show(); 

    } 

怎麼辦?

+0

Offtopic:你在這條線的無效的雙引號:'行[field.name .index] = obj [field.name.name]「;'(倒數第二個字符) – LinkinTED

+0

執行'stringify'然後解析它的目的是什麼? – Utkarsh

+0

請粘貼Json對象 – bpavlov

回答

0

如果你的JSON看起來是這樣的:

{"package": "package-a", "priority": "1", "dcnRefNo": "DCN/Duraline/001", "name": "abc"} 

你可以使用此代碼示例玩:

var data = [{"package": "package-a", "priority": "1", "dcnRefNo": "DCN/Duraline/001", "name": "abc"}]; 
function printJson(data){ 
      var strJson = JSON.stringify(data); 
      var dataJson = JSON.parse(strJson); 
      $.each(dataJson, function(index, obj) { 
       console.log(obj.dcnRefNo); 
       console.log(obj.name); 
       console.log(obj.package); 
       console.log(obj.priority); 
     }); 
} 
+0

該值未獲取。 – anukuls

+0

請參閱上面的評論。 – bpavlov