我似乎无法从引导站点获取示例下拉列表。
我脑子里有以下内容
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css" rel="stylesheet">
<link href="https://maxcdn.bootstrapcdn.com/font-awesome/4.3.0/css/font-awesome.min.css" rel="stylesheet">
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/js/bootstrap.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
这是下拉列表:
<div class="dropdown">
<button class="btn btn-default dropdown-toggle" type="button" id="dropdownMenu1" data-toggle="dropdown" aria-haspopup="true" aria-expanded="true">
Dropdown
<span class="caret"></span>
</button>
<ul class="dropdown-menu" aria-labelledby="dropdownMenu1">
<li><a href="#">Action</a></li>
<li><a href="#">Another action</a></li>
<li><a href="#">Something else here</a></li>
<li><a href="#">Separated link</a></li>
</ul>
</div>
我做错了吗?
答案 0 :(得分:3)
在jQuery
之前加入Bootstrap JS
因为Twitter Bootstrap需要jQuery库才能运行。
<link href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/css/bootstrap.min.css" rel="stylesheet">
<link href="https://maxcdn.bootstrapcdn.com/font-awesome/4.3.0/css/font-awesome.min.css" rel="stylesheet">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.5/js/bootstrap.min.js"></script>
<div class="dropdown">
<button class="btn btn-default dropdown-toggle" type="button" id="dropdownMenu1" data-toggle="dropdown" aria-haspopup="true" aria-expanded="true">
Dropdown
<span class="caret"></span>
</button>
<ul class="dropdown-menu" aria-labelledby="dropdownMenu1">
<li><a href="#">Action</a>
</li>
<li><a href="#">Another action</a>
</li>
<li><a href="#">Something else here</a>
</li>
<li><a href="#">Separated link</a>
</li>
</ul>
</div>