To the contributers so far... You guys rock, thanks again and again. Quick coding question... I wonderful and helpfull script posted by Orio but Im having trouble getting the image to show I have commented the problem area. The link to my image is in a mySQL table and instead of showing the image its showing the code. What the deal here? <?php if(!isset($_POST['submit'])) { die("<html><body><form action=\"".$_SERVER['PHP_SELF']."\" method=\"POST\"> <select name=\"cat\"> <option value=\"Holiday\">Holiday</option> <option value=\"Animals\">Animals</option> </select> <input type=\"submit\" name=\"submit\" value=\"Search!\"> </form> </body> </html>"); } $qry = "SELECT itemName,itemPrice,itemSmall FROM `products` WHERE itemCat='".$_POST['cat']."'"; $result = mysql_query($qry) or die(mysql_error()); echo "<table border=\"1\"><tr><td>Name</td><td>Price</td><td>Thumbnail</td></tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>".$row['itemName']."</td>"; echo "<td>".$row['itemPrice']."</td>"; echo "<td><img src=\"".$row['itemSmall']." /></td>"; \\\itemSmall is the location of my thumbnail on the server echo "</tr>"; } echo "</table>";