我是php和html的新手。我需要从mySQL中检索数据,我帮忙完成了。现在,我需要数据是超链接,点击时,它将被重定向到一个新页面,显示我有关数据的详细信息
<!DOCTYPE HTML>
<html>
<head>
<title>Static IP Address Management Portal</title>
<meta name="description" content="website description" />
<meta name="keywords" content="website keywords, website keywords" />
<meta http-equiv="content-type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="css/style.css" />
<link rel="stylesheet" href="css/skeleton.css">
<link rel="stylesheet" href="css/layout.css">
<!-- modernizr enables HTML5 elements and feature detects -->
<script type="text/javascript" src="js/modernizr-1.5.min.js"></script>
</head>
<body>
<div class="container">
<div class="sixteen columns">
<div id="logo">
<div id="logo_text">
<!-- class="logo_colour", allows you to change the colour of the text -->
<img class="logo" src="images/nyp.png" alt="NYP"/>
<font size ="3.5"><h1>Static IP Address Management Portal </h1></font>
<h2 class="text"><font color="red">View Pool</font> </h2>
</div>
</div>
</div>
</div><!-- End container -->
<div class="container">
<font size="4" >
<nav>
<ul class="sf-menu" id="nav">
<li><a href="Admin.php">Home</a></li>
<li><a href="CreateIPPool.php">Create Pool</a></li>
<li class="selected"><a href="ViewIPPool.php">View Pool</a></li>
<li><a href="Request.php">Request</a></li>
<li><a href="Logout.php">Logout</a></li>
</ul>
</nav>
</font>
</div><!-- End container -->
<div class="container">
<font size="4" >
<div class="sixteen columns">
<div id="site_content">
<div id="content">
<div class="center">
<?php
$servername = "localhost";
$username = "root";
$password = "123456";
$dbname = "ipPool";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT PoolName FROM PoolName";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
echo "<table><tr><th>PoolName</th></tr>";
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["PoolName"]. "</td></tr>";
}
echo "</table>";
} else {
echo "0 results";
}
$conn->close();
?>
</div>
</div>
</div>
</div>
</font>
</div><!-- End container -->
<div class="container">
<font size="4" >
<footer>
<p>Copyright © 2015 Nanyang Polytechnic, School of Information Technology. </p>
</footer>
</font>
</div><!-- End container -->
<p> </p>
<!-- javascript at the bottom for fast page loading -->
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="js/jquery.easing-sooper.js"></script>
<script type="text/javascript" src="js/jquery.sooperfish.js"></script>
<script type="text/javascript">
$(document).ready(function() {
$('ul.sf-menu').sooperfish();
});
</script>
</body>
</html>
答案 0 :(得分:1)
尝试更新
echo "<tr><td>" . $row["PoolName"]. "</td></tr>";
类似
echo "<tr><td><a href='your desired url Goes here'>" . $row["PoolName"]. "</a></td></tr>";
答案 1 :(得分:0)
因此 您必须存储网站以及数据库 。
假设您的数据库包含列名website
然后 查询 应为
$sql = "SELECT PoolName,website FROM PoolName";
所以while
循环
echo "<tr><td><a href='".$row["website"]."'>". $row["PoolName"]. "</a></td></tr>";
答案 2 :(得分:-1)
试试这个,您需要更改超链接以导航到其他地方。
echo "<tr><td><a href='google.com'>". $row["PoolName"]. "</a></td></tr>";