这里的正确人员是我的大屁股问题,我只是一个乞丐和所有这一切,所以它可能只是一个愚蠢而简单的事情,但是如何从这段代码
<?php
session_start();
if(!isset($_SESSION['password']))
{
header('Location:index.php');
exit();
if ($_SESSION['logged_in']= 0)
{
header('Location:index.php');
exit();
}
}
?>
<?php include 'includes/header.php';?>
<div class="page_menu_box">
<ul>
<li><a href="userprofile.php">USER PROFILE</a></li>
<li><a href="aboutus.php">ABOUT US</a></li>
</ul>
</div>
<div class="page_content">
<table class="viewbook_table">
<tr>
<th>ISBN</th>
<th>AUTHOR</th>
<th>DATE WHEN PUBLISHED</th>
<th>COVER</th>
</tr>
<?php
include("core/database/connect.php");
$connection = @new mysqli('localhost','root','','bs_admin_tools');
$sql = "SELECT * FROM books";
$query = $connection->query( $sql );
while ($row = mysqli_fetch_array($query))
{
echo "<tr>";
echo "<td width = 15%>";
echo $row['ISBN'];
echo "</td>";
echo "<td width = 15%>";
echo $row['author'];
echo "</td>";
echo "<td width = 15%>";
echo $row['datepublished'];
echo "</td>";
//check whether an image is available for the record chosen
if ( $row['imgdir'] )
{
//create a variable which holds data from the images folder
$imageDir = "image/";
//create a second variable which holds the value of the image linked to the record selected
$img = $imageDir . $row['imgdir'];
/*link to the directory*/
echo "<td width ='15%' padding='30%'>";
//display the image
echo "<left><a href='viewdetails'><img src='$img'></a>";
echo "</td></tr>";
}
//echo "<br />"; //display a line break
}//end of the while loop
//release connection from database
mysqli_close($connection);
?>
</table>
</div>
<?php
echo "<div class='page_menu_box'><ul><li>
".$_SESSION['login']."
</ul></li></div>'";
include 'includes/footer.php';?>
我只想基本点击$ img并查看书籍详细信息,换句话说,我将其中一个变量变成一个超链接,每当有人被转移到另一个页面时,我想要一个变量被发送到那个文件我会方便文件引入变量,并使用&#39;从书中选择$ theVariable =&#34; XD&#34;&#39;来显示关于该书的其他狗屎。如果有人能帮助我,我会非常高兴,我会在第二天在墙上蹦蹦跳跳。
我只能在php上做一些事情所以请解决方案只涉及php或html; _;