从数据库使用PHP的超链接数据

我是新的PHP和HTML。我需要从mySQL中检索数据，并且我帮助完成了。现在，我需要的数据是超链接，当点击，它会被重定向到一个新的页面，会显示有关数据从数据库使用PHP的超链接数据

This is the data I retrieve from the database. I need it to be hyperlink and each data to be redirected to their respective php website

<!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 &copy; 2015 Nanyang Polytechnic, School of Information Technology. </p> 
      </footer> 
     </font> 
    </div><!-- End container --> 

    <p>&nbsp;</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>

来源

2015-10-26 WeeHau

-1

试试这个我的细节，你需要改变的超级链接导航到其他地方。

echo "<tr><td><a href='google.com'>". $row["PoolName"]. "</a></td></tr>";

来源

2015-10-26 08:27:21 MuthaFury

哇有人恨我，我想上面的那个人讨厌我。您的信息，我没有downvoted你 – MuthaFury

。 –

我不明白为什么我的回答不准确或错误。 – MuthaFury

尝试更新

echo "<tr><td>" . $row["PoolName"]. "</td></tr>";

要像

echo "<tr><td><a href='your desired url Goes here'>" . $row["PoolName"]. "</a></td></tr>";

来源

2015-10-26 08:29:03 Arpita

让你必须也存储在那里的网站在数据库。

假设你的数据库包含列名website

然后查询应该

$sql = "SELECT PoolName,website FROM PoolName";

所以while环

echo "<tr><td><a href='".$row["website"]."'>". $row["PoolName"]. "</a></td></tr>";

来源

2015-10-26 08:30:47

他要求显示一些信息..如果他有10行，他寒create创建10页?? –

@NiranjanNRaju好打听。检查他的问题上的图像。只有两个。所以没有伤害。 –

我同意你的看法，如果它超过100，最好给予灵活的回答，以便他不会再回来。 –

从数据库使用PHP的超链接数据

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