Using PHP to display button with hyperlink or greyed out button if no URL in database

Posted by Diane on Stack Overflow See other posts from Stack Overflow or by Diane
Published on 2010-04-11T02:54:58Z Indexed on 2010/04/11 3:03 UTC
Read the original article Hit count: 327

I've got a webpage that I'm working on where you click on a letter or category and it displays records matching that query from a database. One of the things I want to display is a hyperlinked button that says "Website" if the database record contains a URL in the 'URL' field, and if there is no value in that field, it will display a greyed out version of that button.

I tried using an if...else statement, but was absolutely unable to get the syntax correct trying to get php to call up the 'URL' value in the middle of an "echo "

So here's what I did:

<?php if($row_rsmemalpha['URL'] != NULL) ?><a
 href="http://<?php echo ($row_rsmemalpha['URL']);?>"><target
 ="_blank"><img src="web_button_on.gif"
 alt="Website" border="0" height="18" width="103" /></target></a>

<?php if($row_rsmemalpha['URL'] == NULL)
echo "<img src=\"web_button_off.gif\" alt=\"No Website Available\" height=\"18\" width=\"103\" />";
?>

If there is a URL available it shows the button properly. But if there isn't a URL in the database it shows both buttons.

I have spent a few days studying examples and tutorials on the web, but haven't found too much that helps. The buttons were completely non-functional when I started, so I'm pretty proud of getting this far with it! I've just run out of time and patience for more trial-and-error experimenting.

Any help is appreciated...

Diane

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