Try:
<?php if (!empty($user['social_media']['linkedin'])):?>
<td valign="bottom" align="center">
<?php echo $html->link($html->image('linkedin.gif', array ('alt' =>
'LinkedIn' , 'align' => 'bottom')), "{$user['social_media']
['linkedin']}", array(), null, false); ?>
<?php echo $user['social_media']['linkedin']; ?>
</td>
<?php endif; ?>
Essentially store the value you wish to act as a URL in a variable and
call that variable. Also remember PHP treats single quotes as strings,
and will not search for variables inside them; whereas it will do so
in double quotes. If it's an array you're calling in double quotes,
remember to use curly brackets around it.
On 1 June, 20:27, bobbiloo <b...@liu.com> wrote:
> In this example:
>
> <?php if (!empty($user['social_media']['linkedin'])):?>
> <td valign="bottom" align="center">
> <?php echo $html->link($html->image('linkedin.gif', array ('alt' =>
> 'LinkedIn' , 'align' => 'bottom')), ' ', array(), null, false); ?>
> <?php echo $user['social_media']['linkedin']; ?></td>
> <?php endif; ?>
>
> 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 between the single quotes, I get a parse
> error.
>
> How can I retrieve this table field to be the URL in my $html->link
> statement?
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