php question: do not display if parameter is empty

[pieter-jan]

In a small module i use the following to display a link based on a parameter:

<a href="<?php echo $params->get('block_' . $i . '_urlsite'); ?>" class="sitelink">Website</a>

If the field is left empty, it of course still shows the link (making it href="/")

How can i make it so that if the corresponding field in the backend is left empty, the link as a whole is not shown?

[JTema]

try this way

Code: Select all

<?php 
$yourparamater=$params->get('yourparamater')
if($yourparamater!= null){?>
			<a href="<?php echo $params->get('block_' . $i . '_urlsite'); ?>" class="sitelink">Website</a> 
			<?php }else{?>
Leave here blank
<?php }?>

[pieter-jan]

Parse error: syntax error, unexpected T_IF in /home/hofrenesse/domains/hofvanrenesse.nl/public_html/modules/mod_hvr_stands/tmpl/default.php on line 110

[pieter-jan]

i realize that the php version would help: 5.3.29

[zomtec]

missing ; at the end of your line 109

[pieter-jan]

Thats it, you guys are livesavers. thnx!!!

[JTema]

Yees zomtec is right ; should be at the end of that line

Code: Select all

$yourparamater=$params->get('yourparamater') ;