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Displaying image in php/mysql


depojones

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Hello all,

 

I can select all images from my database and image directories. The problem am left with is the arrangement. I want each picture to be in this order

 

Picture 1          picture2            Picture 3

description        description          description

 

Instead, they are all on a single line. Here is my code:

 

<?php 
require_once 'includes/config.php';
$rowsPerPage =12;
$pageNum = 1;
if(isset($_GET['page']))
{
    $pageNum = $_GET['page'];
}

$offset = ($pageNum - 1) * $rowsPerPage;

$query = " SELECT * FROM lets_products where status='0' order by prod_id DESC" . " LIMIT $offset, $rowsPerPage";
$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result))
{

   echo "
   <a href=\"javascript:popup('$productsdir$row[prod_image]');\">
   <img src=$productsdir$row[prod_image] width=\"100\" height=\"80\" border=\"0\">
   </a><br>
    <strong>$row[prod_name]</strong>
<br>"; ?>
            <a href="javascript:popup('prod_desc.php?prod_id=<?php echo $row['prod_id']; ?>');">click here for more..</a>
            <?php echo "
<br><br>"; 

}
echo "<br><br>";
$query   = "SELECT COUNT(prod_name) AS numrows FROM lets_products where status='0'";
$result  = mysql_query($query) or die(mysql_error());
$row     = mysql_fetch_array($result, MYSQL_ASSOC);
$numrows = $row['numrows'];

$maxPage = ceil($numrows/$rowsPerPage);

$self = $_SERVER['PHP_SELF'];
$nav  = '';
for($page = 1; $page <= $maxPage; $page++)
{
   if ($page == $pageNum)
   {
      $nav .= " $page "; // no need to create a link to current page
   }
   else
   {
      $nav .= "<a href=\"$self?page=$page\">$page</a>";
   } 
}

// creating previous and next link plus the link to go straight to the first and last page

if ($pageNum > 1)
{
   $page  = $pageNum - 1;
   $prev  = " <a href=\"$self?page=$page\">Prev</a> ";

   $first = " <a href=\"$self?page=1\">First</a> ";
} 
else
{
   $prev  = ' '; // we're on page one, don't print previous link
   $first = ' '; // nor the first page link
}

if ($pageNum < $maxPage)
{
   $page = $pageNum + 1;
   $next = " <a href=\"$self?page=$page\">-<img src=\"images/orange_right.gif\"></a> ";


   $last = " <a href=\"$self?page=$maxPage\">-<img src=\"images/orange_right.gif\"><img src=\"images/orange_right.gif\"></a> ";
} 
else
{
   $next = ' '; // we're on the last page, don't print next link
   $last = ' '; // nor the last page link
}

// print the navigation link
echo $first . $prev . "Page $pageNum of $maxPage " . $next . $last;

?>

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https://forums.phpfreaks.com/topic/155604-displaying-image-in-phpmysql/
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because this is not a PHP question, this is an HTML question...

 

but, since you're here.. Just do something like this:

 

<style type="text/css">
div.pice {
	display: block;
	width: 100px;
	float: left;
}
div.pice img {
	display: block;
	width: 100px;
	height: 80px;
}
div.pice div {
	clear: both;
	display: block;
	font-size: 11px;
	font-family: Tahoma;
}
</style>
<div class="pice">
<img src="img_url" />
<div>description</div>
</div><div class="pice">
<img src="img_url" />
<div>description</div>
</div><div class="pice">
<img src="img_url" />
<div>description</div>
</div>

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