如何訪問

Question

我有一個JSON字符串從我的數據庫表來JSON字符串的屬性，但它具有空的根元素名稱

{"": [ 
    {"ID":18,"MenuName":"dsadasdasd","IsActive":"InActive"}, 
    {"ID":17,"MenuName":"Karachi","IsActive":"Active"}, 
    {"ID":2,"MenuName":"User Management","IsActive":"Active"}, 
    {"ID":1,"MenuName":"Home","IsActive":"Active"} 
]} 

我試圖與訪問下面的jquery ajax調用方法這個JSON

function Get_DataTable() { 
    $.ajax({ 
     url: "GridView_JqueryFunctionality.aspx/CallDataTable_EmptyRootName", 
     type: "POST", 
     data: '{}', 
     contentType: "application/json; charset=utf-8", 
     dataType: "json", 
     success: function (data) { 
      alert(data.d) // showing json is fine 
      var MyData = $.parseJSON(data.d); 
      for (i = 0; i < Object.keys(MyData).length; i++) { 
       alert(MyData[i].ID + ' : ' + MyData[i].MenuName); 
      } 
     } 
    }); 
} 

我WEBMETHOD

[WebMethod(true)] 
public static string CallDataTable_EmptyRootName(){  
    List<Category> Categories = new List<Category>(); 
    clsMenu objMenu = new clsMenu(); 
    DataTable dt = new DataTable(); 
    objMenu.GetAllMenu(dt); 
    if (dt.Rows.Count > 0){ 
     string jsonString = ConversionExtension.DataTabelToJson(dt);    
     return jsonString.ToString();   
    }else{ 
     return ""; 
    } 
} 

它給我undefined：undefined ...請幫助我。我現在被卡住了

Answer 1

試試這個：

function Get_DataTable() { 
    $.ajax({ 
     url: "GridView_JqueryFunctionality.aspx/CallDataTable_EmptyRootName", 
     type: "POST", 
     data: '{}', 
     contentType: "application/json; charset=utf-8", 
     dataType: "json", 
     success: function (data) { 
      alert(data.d) // showing json is fine 
      var MyData = $.parseJSON(data.d)[""]; 

      for (i = 0; i < MyData.length; i++) { 
       alert(MyData[i].ID + ' : ' + MyData[i].MenuName); 
      } 
     } 
    }); 
} 

Answer 2

假設你有一個變量名稱爲你的對象，你可以引用它作爲鍵的空字符串。

var a = { 
    "": [ 
    { 
     "ID": 18, 
     "MenuName": "dsadasdasd", 
     "IsActive": "InActive" 
    }, 
    { 
     "ID": 17, 
     "MenuName": "Karachi", 
     "IsActive": "Active" 
    }, 
    { 
     "ID": 2, 
     "MenuName": "User Management", 
     "IsActive": "Active" 
    }, 
    { 
     "ID": 1, 
     "MenuName": "Home", 
     "IsActive": "Active" 
    } 
    ] 
} 
console.log(a[""]); 

嘗試運行上面的代碼通知，我們仍然可以訪問對象元素，儘管有一個空的根名稱。

Answer 3

您必須使用括號訪問根對象中的no-name屬性，然後迭代項目數組。

success: function(response) { 
    var data = $.parseJSON(response.d); 
    var list = data[""]; 
    list.forEach(function(item, i) { 
    console.log(item); 
    console.log(item.ID, ' -> ', item.MenuName); 
    }); 
}