Using PHP to display button with hyperlink or greyed out button if no URL in database
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by Diane
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Published on 2010-04-11T02:54:58Z
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2010/04/11
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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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