如何通过子节点递归并输出为json?

时间:2016-08-15 16:37:52

标签: javascript jquery json

我已经嵌套了包含儿童的无序列表,我试图将列表中的所有数据存储到JSON格式数据中:

以下是我的HTML代码:

<nav id="nav">
    <ul class="header-top-nav">
        <li>
            <a href="index.html">Home</a>
            <ul>
                <li><a href="subhome1.html">Home 1</a>
                    <ul>
                        <li><a href="subsubhome1.html">Sub Home</a></li>
                    </ul>
                </li>
                <li><a href="subhome2.html">Home 2</a></li>
                <li><a href="subhome3.html">Home 3</a></li>
            </ul>
        </li>
    </ul>
</nav>

我的JQuery:

content = [];
$('nav > ul').find('li').each( function () {

        pages = {};
        function menu() {
            pages['pagelink'] = $(this).children('a').attr('href');
            pages['pagename'] = $(this).children('a').text();
            var submenucheck = $(this).children('ul').length;
            if(submenucheck){
                pages['submenu'] = [];
                menu();
            }
            else{
                pages['submenu'] = "NULL";
            }
            contents.push(pages);
        }menu();

});

我希望数据格式如下:

{
  "pages": [
    {
      "pagelink": "index.html",
      "pagename": "Home",
      "submenu": [
        {
          "pagelink": "subhome1.html",
          "pagename": "Home 1",
          "submenu": [
            {
              "pagelink": "subsubhome1.html",
              "pagename": "Sub Home",
              "submenu": "NULL"
            }
          ]
        },
        {
          "pagelink": "subhome2.html",
          "pagename": "Home 2",
          "submenu": "NULL"
        },
        {
          "pagelink": "subhome3.html",
          "pagename": "Home 3",
          "submenu": "NULL"
        }
      ]
    }
  ]
}

请帮助我实现这一目标。

0 个答案:

没有答案