jquerymobile面板交叉页面
我想为每一页制作相同的面板菜单
因为面板菜单是每个页面的相同内容
我该怎么办?或任何建议?
我的示例代码如下
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1, user-scalable=no">
<link href="https://code.jquery.com/mobile/1.4.5/jquery.mobile-1.4.5.min.css" rel="stylesheet" type="text/css">
<title></title>
</head>
<body>
<div data-role="page" id="status">
<div data-role="header" data-theme="b">
<h1>Access Point Status</h1>
<a class="ui-btn" data-rel="back" style="background:none;">back</a>
<a class="ui-btn ui-btn-right" style="background:none;" href="#left-menu">menu</a>
</div>
<div id="left-menu" data-role="panel" data-position="left">
<a class="ui-btn" href="#status">STATUS</a>
<a class="ui-btn" href="#A">A</a>
</div>
<div data-role="content" class="ui-content" data-theme="e">
</div>
</div>
</div>
<div data-role="page" id="A">
<div data-role="header" data-theme="b">
<h1>PAGE A</h1>
<a class="ui-btn " data-rel="back" style="background:none;">back</a>
<a class="ui-btn ui-btn-right" style="background:none;" href="#left-a">menu</a>
</div>
<div id="left-a" data-role="panel" data-position="left">
<a class="ui-btn" href="#status">STATUS</a>
<a class="ui-btn" href="#A">A</a>
</div>
<div data-role="content" class="ui-content" data-theme="e">
</div>
</div>
<script src="http://code.jquery.com/jquery-1.12.4.min.js"></script>
<script src="https://code.jquery.com/mobile/1.4.5/jquery.mobile-1.4.5.min.js"></script>
</html>
&#13;
答案 0 :(得分:0)
在body标签和外页div中创建一个面板,如下所示
paste
然后在第一页的pagecreate事件中添加此
<body>
<div data-role="panel" id="myPanel" data-display="overlay" data-position="right" data-theme="b">
<ul data-role="listview">
<li data-icon="false">
<a href="#" data-transition="pop" data-rel="close">
Home
</a>
</li>
<li data-icon="false">
<a href="#login"data-rel="close">Logout</a>
</li>
</ul>
</div>
<div data-role="page" id="login">
..... page ....
</div>
.... other pages ....
</body>