<?php echo "<img src='/album/img/".$photoFileName[2]." width="90" height="70" border='0' / >"; ?> Should be: <?php echo "<img src='/album/img/" . $photoFileName[2] . " width=\"90\" height=\"70\" border='0' / >"; ?> Or <?php echo "<img src='/album/img/" . $photoFileName[2] . " width='90' height='70' border='0' / >"; ?> Or <?php echo '<img src="/album/img/' . $photoFileName[2] . ' width="90" height="70" border="0" / >'; ?> And the third one is the best. -----Original Message----- From: elk dolk [mailto:elkdolk@xxxxxxxxx] Sent: Friday, March 30, 2007 7:32 PM To: php-db@xxxxxxxxxxxxx Subject: width& height it might be a stupid question but I think it would be the last one! I am trying to define the height and width of picture in the following line: "<img src='/album/img/".$photoFileName[2]." / >"; ?></td> when I put it like this: <table width="50%" border="0" cellspacing="3" cellpadding="0"> <tr> <td width="90" height="70"><?php echo "<img src='/album/img/".$photoFileName[2]." width="90" height="70" border='0' / >"; ?></td> </tr> <tr> <td width="90" height="70"> </td> <td width="90" height="70"> </td> </tr> </table> <p> </p> <p> I have this error: PHP Parse error: syntax error, unexpected T_LNUMBER, expecting ',' or ';' in C:\Inetpub\wwwroot\album\2dimArray2.php on line 44 please comment! --------------------------------- Bored stiff? Loosen up... Download and play hundreds of games for free on Yahoo! Games. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
