2

我正在尝试使用 twitter bootstrap 构建动态菜单。这个想法是加载一个包含菜单项的 json 文件：

// test.json

{
"children": [
    {
    "text": "Item1",
    "children": [
        {"text": "Item11", "link": "/link11"},
        {"text": "Item12", "link":"/link12"},
        {"text": "Item13", "children": [
            {"text": "Item131", "link":"/link131"},
            {"text": "Item132", "link":"/link132"}
        ]},
        {"text": "Item14", "link":"/link13"}
    ]},
    {
    "text": "Item2",
    "children": [
        {"text": "Item21", "link":"/link21"},
        {"text": "Item22", "link":"/link22"}
    ]},
    {
    "text": "Item3"
    }
]
}

这可以在 Twitter 引导程序中静态呈现：

// index.html

<!DOCTYPE html>
<html lang="en">
<head>
   <title>Test</title>
   <meta name="viewport" content="width=device-width, initial-scale=1.0">
   <link href="bootstrap/css/bootstrap.min.css" rel="stylesheet" media="screen">
</head>
<body>
<div class="navbar">
    <div class="navbar-inner" id ="navbar-inner-main">
        <div class="container">
            <div class="nav-collapse collapse navbar-responsive-collapse">
                <ul class="nav">
                    <li class="dropdown">
                        <a class="dropdown-toggle" data-toggle="dropdown" href="#">Item1</a>
                        <ul class="dropdown-menu">
                            <li>
                                <a href="/link11">Item11</a>
                            </li>
                            <li>
                                <a href="/link12">Item12</a>
                            </li>
                            <li class="dropdown-submenu">
                                <a class="dropdown-toggle" data-toggle="dropdown" href="#">Item13</a>
                                <ul class="dropdown-menu">
                                    <li>
                                        <a href="/link131">Item131</a>
                                    </li>
                                    <li>
                                        <a href="/link132">Item132</a>
                                    </li>   
                                </ul>
                            </li>
                        </ul>
                    </li>
                    <li class="dropdown">
                        <a class="dropdown-toggle" data-toggle="dropdown" href="#">Item2</a>
                        <ul class="dropdown-menu">
                            <li>
                                <a href="/link21">Item21</a>
                            </li>
                            <li>
                                <a href="/link22">Item22</a>
                            </li>   
                        </ul>
                    </li>
                    <li>
                        <a href="#">Item3</a>
                    </li>
                </ul>
            </div>
        </div>
    </div>
</div>     
<script src="http://code.jquery.com/jquery.js"></script>
<script src="bootstrap/js/bootstrap.min.js"></script>

我想在 index.html 中添加这样的函数：

<script type="text/javascript">
    $(document).ready(function () {
      $.getJSON('ajax/test.json', function(data) {
      // get json, parse it and append to .nav
      });     
});
</script>

我认为将 json 解析为无序列表的最佳方法是使用模板。我尝试了纯 和小胡子，但没有成功。问题是由于下拉子菜单类使递归方法不一致。

你有什么建议吗？

提前致谢

4

1 回答 1

2

我有类似的问题。这是解决方案：

我的 JSON

[{
"name": "Cars",
"url": "#",
"parent_id": "-1",
"id": "1"
 }, {
"name": "Toyota",
"url": "#",
"parent_id": "1",
"id": "2"
    }, {
"name": "Corolla",
"url": "#",
"parent_id": "2",
"id": "3"
 }, {
"name": "2013",
"url": "#",
"parent_id": "3",
"id": "4"
    },
...

这是使用引导类生成下拉列表的 JS

//load JSON file
$.getJSON('menu.json', function(data) {
//build menu
var builddata = function () {
var source = [];
var items = [];
for (i = 0; i < data.length; i++) {
    var item = data[i];
    var label = item["name"];
    var parentid = item["parent_id"];
    var id = item["id"];
    var url = item["url"];

    if (items[parentid]) {
        var item = { parentid: parentid, label: label, url: url, item: item };
        if (!items[parentid].items) {
            items[parentid].items = [];
        }
        items[parentid].items[items[parentid].items.length] = item;
        items[id] = item;
    }
    else {
        items[id] = { parentid: parentid, label: label, url: url, item: item };
        source[id] = items[id];
    }
}
return source;
}

var buildUL = function (parent, items) {
$.each(items, function () {
    if (this.label) {
        var li = $("<li class='js-menu'>" + "<a href='"+ this.url +"'>" + this.label + "      </a></li>");
        li.appendTo(parent);
        if (this.items && this.items.length > 0) {
            var ul = $("<ul class='dropdown-menu js-menu'></ul>");
            ul.appendTo(li);
            buildUL(ul, this.items);
        }
    }
});
}
var source = builddata();
var ul = $(".json-menu");
ul.appendTo(".json-menu");
buildUL(ul, source);
//add bootstrap classes
if ($(".json-menu>li:has(ul.js-menu)"))
{
$(".json-menu>li.js-menu").addClass('dropdown-submenu');
}
if ($(".json-menu>li>ul.js-menu>li:has(> ul.js-menu)"))
{
$(".json-menu>li>ul.js-menu li ").addClass('dropdown-submenu');
}
$("ul.js-menu").find("li:not(:has(> ul.js-menu))").removeClass("dropdown-submenu");
});

所以你只需要在'dropdown' bootsrap 类附近添加'json-menu' 类

<ul class="dropdown-menu json-menu">
</ul>

你可以在这里看到活的例子

于 2013-06-07T13:55:32.613 回答