How to call site name in PHP block?

Question

I have a block in my footer that I currently use to display a copyright message using PHP, like so:

Copyright © <?php echo date("Y"); ?>

I'd like to add the site name to that code snippet, dynamically instead of typing the name. How can I do so?

So the end result would be:

Copyright © <?php echo date("Y"); ?> [[[php code to display site name goes here]]]

Thanks!

Answer 1

You can use variable_get() to get the site name, but you shouldn't use it directly in the template file or you could break caching.

Instead, define a preprocess function in your theme's template.php file:

function MYTHEME_preprocess_page(&$vars) {
  $vars['site_name'] = variable_get('site_name', 'Default');
}

Then you can use this in the template file:

Copyright © <?php echo date("Y"); ?> <?php echo $site_name; ?>

Before doing that, though, check whether you already have $site_name available in page.tpl.php. Some themes add it for you, some don't.

Answer 2

As an alternative to creating a pre-process function in your template, you can print the value of the site name variable in a block using this php snippet in your block

<?php
$site_name = variable_get('site_name');
echo $site_name;
?>

So, to print a copyright notice that includes your site name, it would look something like:

©copyright <?php echo date('Y');?><?php $site_name = variable_get('site_name'); echo "&nbsp"; echo $site_name; echo", "; ?> All Rights Reserved.