You're trying to use HTML within your PHP code, so PHP sees this as an unexpected variable/string. Either use echo
for this, or close the PHP statement, and then write your HTML.
Either:
<div id="social_icon">
<?php if(isset($fburl)){ ?>
<a href="<?php echo $options['fburl']; ?>">
<img src="http://farm6.staticflickr.com/5534/9135106179_43deba3d15_o.png" width="30" height="30" />
</a>
<?php }else{
//dont show anything
} ?>
</div>
Or:
<div id="social_icon">
<?php if (isset($fburl)){
echo '<a href="'.$options['fburl'].'"><img src="http://farm6.staticflickr.com/5534/9135106179_43deba3d15_o.png" width="30" height="30" /></a>';
}else{
//dont show anything
} ?>
</div>
Edit
Actually, I would assume it's not outputting anything because your if statement is checking for $fburl
whereas you're echo
ing the link as $options['fburl']
. If the facebook url is located at $options['fburl']
, try:
<div id="social_icon">
<?php if(isset($options['fburl'])){ ?>
<a href="<?php echo $options['fburl']; ?>">
<img src="http://farm6.staticflickr.com/5534/9135106179_43deba3d15_o.png" width="30" height="30" />
</a>
<?php }else{
//dont show anything
} ?>
</div>
Edit 2
If the options are set but don't contain a link, you will also need check for that:
<div id="social_icon">
<?php if(isset($options['fburl']) && !empty($options['fburl'])){ ?>
<a href="<?php echo $options['fburl']; ?>">
<img src="http://farm6.staticflickr.com/5534/9135106179_43deba3d15_o.png" width="30" height="30" />
</a>
<?php }else{
//dont show anything
} ?>
</div>