Jquery 如何附加嵌套的ul

我的Sqlite数据库数据：-

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

在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>'
}

for（var i=0；i

请帮助我，如果我现在不知道有多少级

在一个级别上，我做到了，而且

如何仅针对该级别内的两个或三个子级别执行操作hi i设置一个，其中添加嵌套的

ul

，并提供数据。 你需要做些改变我从虚拟源中得到的数据如下

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;

而不是这个

var food_Id = results.rows.item(i).ID ;
var food_PId = results.rows.item(i).ParentId;
var food_Item = results.rows.item(i).Item;

---

你的问题不太清楚。请详细说明你真正想要什么？你想在

JQueryMobile列表视图中使用它吗？@bluanger是的，我想在JQueryMobile列表视图中使用style@AtulDhanuka我添加了答案检查，如果您需要
以简单样式而不是
JQueryMobile
样式列出
，请删除
data role=“listview”属性
谢谢你的回答最后我得到了答案，也请喜欢我的问题