while($row = mysql_fetch_array($result,MYSQL_ASSOC))
 {
 $image= $row[image];
 ?>
 <tr>
 <td height="24"> <div align="center"> <img src="<?php echo $image; ?>" > </div></td>
 <td><div align="center"><? $row[title]; ?></div></td>
 <td><div align="center"><? $row[startPrice]; ?></div></td>
 </tr>
 <?
 }
 ?>

Ive got a database linked to this table, it displays the first field(image) fine, but doesnt show the next two entries (title and startPrice), any ideas why?

it displays the php code in the correct place in dreamweaver but when i upload to my site and try to view it, only the picture shows up?!

any ideas whats my problem will be appreciated!

NIC OR RICH I NEED YOU!!!hehe!

CHEERS
ASH

__________________
My new web design domain
www.askmultimedia.co.uk

lol, you've neglected to echo the value..

Try
<?php echo $row['title']; ?>
<?php echo $row['startPrice']; ?>
;)

Regards,
Rich

::::::::::::::::::::::::::::::::::::::::::
The Spicy Peanut Project
http://www.spicypeanut.net
::::::::::::::::::::::::::::::::::::::::::

oh my word, i feel utterly stupid!!haha

i think it must be the hours of programming ive done today has pickled my brain!!

cheers rich,
ASH:D

It's not a huge error. PHP uses a "shortcut" notation to echoing values. If you would've used an = sign, it would've worked.

Personally, I HATE this shortcut notation. But here it is anyway:

<?=$var;?>

More examples:
<?= $row['title']; ?>
<?= $row['startPrice']; ?>


Look at the echo docs for more info:
  http://www.php.net/echo


Take care,

Nik
http://www.bigaction.org/