2014-03-05 86 views
1

我SQLite數據庫的數據: -如何追加嵌套UL

ID  ParentId  Item 

1   0   Food 

2   1   Indian 

3   2   Full Meal 

4   2   Meal 

5   4   Roti 

6   4   Dal 

7   0   Chinese 

8   7   Soup 

9   8   Momo 

10   9   Noodle 

jQuery中: -

for (var i = 0; i < results.rows.length; i++) { 
    var food_Id = results.rows.item(i).ID ; 
    var food_PId = results.rows.item(i).ParentId; 
    var food_Item = results.rows.item(i).Item; 
     if(food_PId == 0){ 
      sfoodID = food_Id; 
      foodscripts += '<li><a href="#" class="d-main-menu">' 
          +'<img src="images/1.png" class="e-left" />' 
          +'<span class="d-menu-parent-text e-left">'+food_Item+'</span></a>' 
     }else{ 
      foodscripts += '<ul>' 
      for(var j = 0; j < results.rows.length; j++){ 
       var food_Id_s = results.rows.item(j).ID ; 
       var food_PId_s = results.rows.item(j).ParentId ; 
       var food_Item_s = results.rows.item(j).Item; 
        if(sfoodID == food_PId_s){ 
         foodscripts += '<li><a href="">'+food_Item_s+'</a></li>' 
        } 
      } 
      foodscripts += '</ul>' 
     } 
    foodscripts += '</li>' 
} 

請幫助我,如果我不現在多少級有

達到一級我已經做到了

如何做到兩級或三級子級內只有

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你的問題是不是更clear.please闡述你真正想要什麼? –

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你想在'JQueryMobile listview'風格? – Blu

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@BluAngel是的,我想在jqueryMobile列表視圖樣式 –

回答

1

嗨我設置了一個Demo其中添加嵌套ul與您提供的數據。 你必須需要一些改變,我收到來自僞源數據這樣

var dataArray = new Array(); 
var data = new Array();data['ID'] = 1;data['ParentId'] = 0;data['Item'] = 'Food';dataArray.push(data); 
var data = new Array();data['ID'] = 2;data['ParentId'] = 1;data['Item'] = 'Indian';dataArray.push(data); 
var data = new Array();data['ID'] = 3;data['ParentId'] = 2;data['Item'] ='Full Meal';dataArray.push(data); 
var data = new Array();data['ID'] = 4;data['ParentId'] = 2;data['Item'] = 'Meal';dataArray.push(data); 
var data = new Array();data['ID'] = 5;data['ParentId'] = 4;data['Item'] = 'Roti';dataArray.push(data); 
var data = new Array();data['ID'] = 6;data['ParentId'] = 4;data['Item'] ='Dal';dataArray.push(data); 
var data = new Array();data['ID'] = 7;data['ParentId'] = 0;data['Item'] = 'Chinese';dataArray.push(data); 
var data = new Array();data['ID'] = 8;data['ParentId'] = 7;data['Item'] = 'Soup';dataArray.push(data); 
var data = new Array();data['ID'] = 9;data['ParentId'] = 8;data['Item'] ='Momo';dataArray.push(data); 
var data = new Array();data['ID'] = 10;data['ParentId'] = 9;data['Item'] = 'Noodle';dataArray.push(data); 


var food_Id = dataArray[i].ID; 
var food_PId = dataArray[i].ParentId; 
var food_Item = dataArray[i].Item; 

而是這個

var food_Id = results.rows.item(i).ID ; 
var food_PId = results.rows.item(i).ParentId; 
var food_Item = results.rows.item(i).Item; 
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,如果你需要'列表'在簡單的風格不''JQueryMobile'風格,然後刪除'data-role =「listview」屬性「 – Blu

+0

感謝回答最後我得到了答案,並請喜歡我的問題 –

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歡迎它感到高興,它幫助 – Blu