tet3828 Posted October 26, 2006 Share Posted October 26, 2006 Thanks for the responses so far people. I appreacieate you all bearing with me as I reduce the broad nature of my questions by grasping the mysql/PHP concept. Ill admit the last response made very little sense to me prior to my recent trip to the library and some practice coding....Moving right along.here goes:I have a table in mySQL it containsroughly 20 rows or "items" with the following values as columns: NAMEs, CATAGORYs, DESCRIPTIONs , PRICEs and HTTP:// reference to a thumbnailhow should I approach displaying the NAME PRICE AND THUMBNAIL in a borderless table when an item's catagory is selected from a dropdown? Quote Link to comment Share on other sites More sharing options...
Orio Posted October 26, 2006 Share Posted October 26, 2006 [code]<?phpif(!isset($_POST['submit'])){die("<html><body><form action=\"".$_SERVER['PHP_SELF']."\" method=\"POST\"><select name=\"cat\"><option value=\"cat1\">Cat1</option><option value=\"cat2\">Cat2</option></select><input type=\"submit\" name=\"submit\" value=\"Search!\"></form></body></html>");}//Make sure you are connected to the database$qry = "SELECT NAME,PRICE,HTTP:// FROM `Table_Name` WHERE CATAGORY='".$_POST['cat']."'";$result = mysql_query($qry) or die(mysql_error());echo "<table border=\"0\"><tr><td>Name</td><td>Price</td><td>Thumbnail</td></tr>";while($row = mysql_fetch_array($result)){ echo "<tr>"; echo "<td>".$row['NAME']."</td>"; echo "<td>".$row['PRICE']."</td>"; echo "<td><img src=\"".$row['HTTP://']." /></td>"; echo "</tr>";}echo "</table>";?>[/code]Orio. Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.