我创建了接受来自 $_GET 方法的数据的PHP文件。
之后,我使用获得的所有数据来创建 HTML 页面。数据没有问题,但在CSS中我无法设置 HTML 元素的样式。除非涉及内联样式,否则它可以工作,但不好维护。
我尝试这样使用,但它不起作用,请帮助
预先感谢
例子.php
<?php
$dataCover = $_GET['dataCover'];
$dataTitle = $_GET['dataTitle'];
$dataTag = $_GET['dataTag'];
$dataDir = $_GET['dataDir'];
$dataYear = $_GET['dataYear'];
$dataCreated = $_GET['dataCreated'];
$dataModified = $_GET['dataModified'];
$userAUID = $_GET['userAUID'];
$galleryID = $_GET['galleryID'];
?>
<!DOCTYPE html>
<html>
<head>
<title></title>
<style type="text/css" media="all">
#container img{
height: 230px;
width: 200px;
}
#container .center{
display: block;
margin: 0 auto;
}
</style>
<script src="../lib/jquery-1.10.2.min.js"></script>
<script src="../lib/jquery.mobile-1.3.1.min.js"></script>
<script src="../lib/se.js"></script>
</head>
<body>
<div data-role ="page" id ="page1">
<div data-role ="header">
<h1> header </h1>
</div>
<div data-role="content">
<div id="container">
<img class="center" src="<?echo $dataCover?>" alt=""/>
<p id="title"><?echo $dataTitle;?></p>
<p id="tag"><?echo $dataTag;?></p>
<p id="created">Created : <?echo $dataCreated?></p>
<p id="modified">modified : <?echo $dataModified?></p>
<a href="http://54.249.251.55/AUgallery" target="_blank" data-url="<? echo $dataCreated ?>" rel="external" data-role="button">View Ebook-Gallery</a>
<a href="<?echo 'http://localhost/webAPP/php/addBookmark.php?userAUID='.$userAUID.'¬eID='.$galleryID?>"data-ajax="false" rel="external" data-role="button">Bookmark</a>
</div>
</div>
</div>
</body>
</html>