The faulty line is:
else { echo '<p><img src=$image width="201"
height="160"></p></td>'; }
$image does not interpolate (surrounded by single quotes)
You must write:
else { echo '<p><img src="{$image}" width="201"
height="160"></p></td>'; }
It is good hygiene to brace variables in somehow complicated statements like
these. Keeps life easier.
HTH
Ignatius
____________________________________________
----- Original Message -----
From: <Bzdpltd@aol.com>
To: <php-db@lists.php.net>
Sent: Tuesday, October 22, 2002 7:15 PM
Subject: Output errors
> Hi there,
>
> I am working to a deadline, and missed something probably simple here.
>
> I have this code, and as I have an if else statement inbetween, it is not
outputting the image name from the database, and rather just how I put the
code in ie. $image.
>
> Could someone see where I am going wrong. Help much appreciated. Code
selection below.
>
> if (mysql_num_rows($sql_result) ==0)
> {
> echo (" \n");
> }
> else {
> while ($row = mysql_fetch_array($sql_result)){
> $title = $row["title"];
> $description = $row["description"];
> $image = $row["image"];
>
> echo ("<tr> \n");
> echo ("<td width=\"72%\" valign=\"top\"><h3>$title</h3> \n");
> echo ("<p>$description</p></td> \n");
> echo ("<td width=\"28%\" valign=\"top\"> \n");
> if ($image =='NULL') { echo ''; }
> else { echo '<p><img src=$image width="201"
height="160"></p></td>'; }
> echo ("</tr> \n");
> }}
>
>
> Thanks
>
> Barry
>
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