Output Errors 2
| From: | Bzdpltd at aol dot com | Date: | Tue, 22 Oct 2002 17:57:17 +0000 |
| Subject: | Output Errors 2 | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-22592@lists.php.net to get a copy of this message | ||
Hi sorry back again.
Thanks for helping me get the code right. Sorry being stupid again, I am having problems with an if
/ else statement.
I am trying the following but it does not seem to work, what the problem is, that I am saying if the
image name field in the database is empty to print nothing, and if it contains data to print that
field name, as the a href is in the code which pulls the image, if the field is empty it should show
nothing, but what happens is the image is shown in the browser and its not blank.
Here is the revised code:
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 '<p></p></td>'; }
else { echo '<p><img src="' . $image . '"
width="201" height="160"></p></td>'; }
echo ("</tr> \n");
}}
If someone can see the problem it would be most helpful.
Best wishes
Barry