Hi this is my first post so apologies if I make any mistakes in the format of my post.
Im trying to display an image from my database.
Ok here is the error im getting when I try to view the page in my browser:
Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\photographers3.php on line 102
And here is my code for the php file:
(line 102 has been highlighted by a comment and the PHP script is in bold)
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="en-US">
<head>
<title>Through The Lens - Homepage</title>
<meta http-equiv="content-type" content="text/html; charset=utf-8" />
<link rel="stylesheet" type="text/css" href="template4.css" />
</head>
<body>
<img src="images/topbanner.gif" height = "200" width = "945" >
<div id="menu">
<ul>
<li><a href="index.php" title="Home"><b><span>Home</span></b></a></li>
<li><a href="#" title="Galleries"><b><span>Galleries</span></b></a></li>
<li><a href="#" title="Equipment"><b><span>Equipment</span></b></a></li>
<li><a href="#" title="Photographers"><b><span>Photographers</span></b></a></li>
<li><a href="#" title="Contact"><b><span>Contact</span></b></a></li>
</ul>
</div>
<div id="wrapper">
<div id="content">
<div id="side1">
<p class="sidebartitle">Visit the Galleries</p>
<img src="images/landscape.jpg" alt="landscape">
<img src="images/nature.jpg" alt="nature">
<img src="images/sport.jpg" alt="sport">
<img src="images/people.jpg" alt="people">
</div>
<div id="main">
<h3>Members Page</h3>
<?php>
$link = mysqli_connect('*****', '******', '*******');
if (!$link)
{
$output = 'Unable to connect to the database server.';
include 'output.html.php';
exit();
}
if (!mysqli_set_charset($link, 'utf8'))
{
$output = 'Unable to set database connection encoding.';
include 'output.html.php';
exit();
}
if (!mysqli_select_db($link, 'kmdb'))
{
$output = 'Unable to locate the members database.';
include 'output.html.php';
exit();
}
$output = '';
include 'output.html.php';
$result = mysqli_query($link, 'SELECT * FROM members');
if (!$result)
{
$error = 'Error fetching data: ' . mysqli_error($link);
include 'error.html.php';
exit();
}
while($info = mysqli_fetch_array( $result ))
{
Echo "<img src=http://localhost/images/".$info['picture'] .">
}
mysqli_free_result($result);
?>
</div>
</div>
</div>
<div id="side2"> //This is line 102
</div>
<div id="footer">
<p class="footer"><b>Through The Lense is a project by Marsdens co. Email : example@example.com</b></p>
</div>
</body>
</html>
Any help would be greatly appreciated.
Thank you.
