使用智能菜单 jquery 打印 json [英] printing json using smart menu jquery

问题描述

 {"menu" :[{ 
"name": "Computers",
 "children": [{
  "name": "Notebook",
  "children": [{
   "name": "Apple"
  }, {
   "name": "Windows"
  }]
 }, {
  "name": "Tablets",
  "children": [{
   "name": "Apple"
  }, {
   "name": "Android"
  }, {
   "name": "Windows"
  }]
 }]
}, {
 "name": "Phones",
 "children": [{
  "name": "Android",
  "children": [{
   "name": "Samsung"
  }, {
   "name": "Nokia"
  }, {
   "name": "Lenovo"
  }]
 }, {
  "name": "Windows Phones",
  "children": [{
   "name": "Microsoft"
  }, {
   "name": "Nokia"
  }]
 }]
}, {
 "name": "Cameras",
 "children": [{
  "name": "Digital",
  "children": [{
   "name": "Nikon"
  }, {
   "name": "Fuji"
  }]
 }, {
  "name": "DSLR",
  "children": [{
   "name": "Canon"
  }, {
   "name": "Nikon"
  }]
 }]
}]
}

我想打印它，这样每个父母都有自己的孩子..这是我的代码

and i want to print it so each parent have its children .. here is my code

$(document).ready(function() {
             $.ajax({
              url: 'menuData.json',
              type: 'get',
              dataType: 'json',
              error: function(data){
               alert("error");
              },
              success: function(data){
     
               var i=0;
               var j=0;
               var n=0;
               var obj=data;
               var json = JSON.stringify(obj);
               var s = JSON.parse(json);
               for( i=0;i<s.menu.length;i++){
            	   $("#main-menu").append(' <li><a href="#">'+ s.menu[i].name +'</a></li>');
	             for( j=0;j<s.menu[i].children.length;j++)
	             { $("#main-menu").append(' <li><a href="#">'+  s.menu[i].children[j].name + '</a></li>');
	             for( n=0;n<s.menu[i].children[j].children.length;n++){
	            	
	               $("#main-menu").append(' <li><a href="#">'+ s.menu[i].children[j].children[n].name +'</a></li>');
	            	 
               }
	           
               }
               }

               $('#main-menu').smartmenus({
               	subMenusSubOffsetX:1,
               	subMenusSubOffsetY: -8
               	
               });
                  }
              });

         });

但结果是这样

请提供任何帮助..在此先感谢..顺便说一句，我是初学者，所以请帮助我..再次感谢

any help please .. thanks in advance .. btw i am beginner so please help me .. thanks again

推荐答案

您的代码在逻辑上是正确的，除非您总是附加到 $("#main-menu").考虑这样做:

Your code is logically correct except you are always appending to the $("#main-menu"). Consider doing this:

...

for( var i=0; i < s.menu.length; i++)
{
    $("#main-menu").append(' <li id="menu-list-' + i + '"><a href="#">' + s.menu[i].name + '</a></li>');

    var list_length = s.menu[i].children.length;
    if (list_length > 0)
        $("#main-menu li#menu-list-" + i).append('<ul></ul>');

    for( var j=0; j < list_length; j++)
    {
        $("#main-menu li#menu-list-" + i + " ul").append(' <li id="menu-list-' + i + '-children-list-' + j + '"><a href="#">'+  s.menu[i].children[j].name + '</a></li>');

        var children_list_length = s.menu[i].children[j].children.length;
        if(children_list_length > 0)
            $("#main-menu li#menu-list-" + i + " ul li#menu-list-" + i + "-children-list-" + j).append("<ul></ul>");
        for( var n=0; n < children_list_length; n++)
        {
            $("#main-menu li#menu-list-" + i + " ul li#menu-list-" + i + "-children-list-" + j + " ul").append(' <li><a href="#">'+ s.menu[i].children[j].children[n].name +'</a></li>');
        }
    }
}
...

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