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[fixed] Error Message:


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I have been adapting this code, that I originally wrote in April, however, I am receiving the error message mentioned above. Could anyone point out my error?

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN""http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"><html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"><head><link rel="stylesheet" type="text/css" href="http://www.equinefocus.co.uk/style/20090405.css"/><meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /><title>Equine Focus Photography</title></head> <body><?php	// 	Connects to your Database  		mysql_connect("*****", "******", "*****") or die(mysql_error());  		mysql_select_db("*****") 		or die ("Unable to connect to MySQL");	// mysql			$per_page = 12;		if (!isset($_GET['page']) || !is_numeric($_GET['page'])) {		$page = 1;		}		else {		$page = (int)$_GET['page'];		}		$from = (($page * $per_page) - $per_page);				$sql = "SELECT * FROM photos WHERE date='2009-05-30' LIMIT $from, $per_page";?><?php		while($info = mysql_fetch_array($data)) {  			// images		$thumb='/photos/'.$info['date'].'/thumbs/'.$info['thumb'];		$url='/photos/'.$info['date'].'/'.$info['url'];		$ref = $info['date'].'-'.$info['ref'];?><?php	$total_results = mysql_fetch_array(mysql_query("SELECT COUNT(*) AS num FROM photos"));	$total_pages = ceil($total_results['num'] / $per_page);	if ($page > 1) {	$prev = ($page - 1);	echo "<a href=\"?page=$prev\"><<  Previous</a> ";	}	for($i = 1; $i <= $total_pages; $i++) {	if ($page == $i) {		echo "$i ";	}	else {		echo "<a href=\"?page=$i\">$i</a> ";	}	}	if ($page < $total_pages) {   $next = ($page + 1);   echo "<a href=\"?page=$next\">Next >></a>";}?>		<div class='img'>				<a href="<?php echo $url;?>"><img src="<?php echo $thumb; ?>"</a>				<div id='refno'><?php echo $ref; ?></div>				<div id='info'><?php echo $info['horse']; ?></div>				<div id='info'><?php echo $info['rider']; ?></div>		</div><?php}  ?></body></html>

The error message would appear to relate to this:

<?php		while($info = mysql_fetch_array($data)) {  			// images		$thumb='/photos/'.$info['date'].'/thumbs/'.$info['thumb'];		$url='/photos/'.$info['date'].'/'.$info['url'];		$ref = $info['date'].'-'.$info['ref'];?>

Any advice greatly appreciated.

Edited by Borderline
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