Try:
link($html->image('linkedin.gif', array ('alt' =>
'LinkedIn' , 'align' => 'bottom')), "{$user['social_media']
['linkedin']}", array(), null, false); ?>
Essentially store the value you wish to act as a URL in a variable and
call that variable. Also remember PHP treats
In this example:
link($html->image('linkedin.gif', array ('alt' =>
'LinkedIn' , 'align' => 'bottom')), ' ', array(), null, false); ?>
I am trying to use a field in my social_media table, entitled
"linkedin," to be the url of my $html->link statement. But if I plug
that php echo line in betwee
