<?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/' . $photoFileName[2] . ' width="90"
height="70" border="0" / >';
?>
And the third one is the best.
-----Original Message-----
From: elk dolk [mailto:[EMAIL PROTECTED]
Sent: Friday, March 30, 2007 7:32 PM
To: php-db@lists.php.net
Subject: [PHP-DB] width& height
it might be a stupid question but I think it would be the last one!
I am trying to define the height and width of picture in the following line:
"<img src='/album/img/".$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!
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