you need to escape double quotes in an echo if you use double quotes as the
wrapper.
bastien
From: elk dolk <[EMAIL PROTECTED]>
To: "php-db@lists.php.net" <php-db@lists.php.net>
Subject: [PHP-DB] width& height
Date: Fri, 30 Mar 2007 04:31:52 -0700 (PDT)
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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