我正在使用HTML / CSS / JS(phoneap,jquery mobile)创建移动应用程序,我正在转换https://build.phonegap.com/apps的代码。
我需要实现这种行为: 点击外部页面上链接的链接后,我需要打开一个选择浏览器的问题。 这是我的代码:
<!DOCTYPE html>
<html>
<head>
<title>Page Title</title>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
<script src="http://code.jquery.com/jquery-1.9.0.js"></script>
<script src="http://code.jquery.com/jquery-migrate-1.2.1.js"></script>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.8.2/jquery.min.js"></script>
<script src="//code.jquery.com/ui/1.11.0/jquery-ui.js"></script>
<script src="http://code.jquery.com/mobile/1.4.3/jquery.mobile-1.4.3.min.js"></script>
<script type="text/javascript" src="js/bootstrap.js"></script>
<script src="https://www.paypalobjects.com/js/external/dg.js" type="text/javascript"></script>
<script type="text/javascript" src="js/datapicker/jquery.datetimepicker.js"></script>
<script src="js/jPayPalCart.js" type="text/javascript"></script>
<script type="text/javascript" src="js/phonegap-1.2.0.js"></script>
<link rel="stylesheet" href="js/datapicker/jquery.datetimepicker.css" />
<link rel="stylesheet" href="css/jquery.mobile-1.1.1.css" />
<script type="text/javascript">
function openDialog() {
window.open('http://apache.org', '_blank', 'location=yes');
}
</script>
<link rel="stylesheet" type="text/css" href="css/jquery-ui-1.10.4.custom.css">
<link rel="stylesheet" type="text/css" href="css/jquery.timepicker.css">
<link rel="stylesheet" type="text/css" href="css/bootstrap.css">
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<!-- Start of first page -->
<div data-role="page" id="findShop">
<div data-role="content">
<div class="container" data-role="content" role="main">
<div class="row main-page">
<div class="col-xs-12 title">
Die vielleicht beste Pizza der Stadt!
<a href="#" onclick="openDialog();">Open in web browser</a>
</div>
</div>
</div>
</div>
</div>
</body>
</html>
有谁知道哪里可能有问题