function pagination($pager)
{
    $uri = sfRouting :: getInstance()->getCurrentInternalUri();
    $html = '';
 
    if ($pager->haveToPaginate())
    {
        $uri .= strstr($uri, '?') ? '&page=' : '?page=';
 
        if ($pager->getPage() != 1)
        {
            $html .= '<li>' . link_to('first', $uri . '1') . '</li>';
            $html .= '<li>' . link_to('previous', $uri . $pager->getPreviousPage()) . '</li>';
        }
 
        foreach ($pager->getLinks() as $page)
        {
            if ($page == $pager->getPage())
                $html .= '<li><strong>' . link_to($page, $uri . $page) . '</strong></li>';
            else
                $html .= '<li>' . link_to($page, $uri . $page) . '</li>';
        }
 
        if ($pager->getPage() != $pager->getLastPage())
        {
            $html .= '<li>' . link_to('next', $uri . $pager->getNextPage()) . '</li>';
            $html .= '<li>' . link_to('last', $uri . $pager->getLastPage()) . '</li>';
        }
 
        $html = '<ul class="pagination">' . $html . '</ul>';
    }
 
    return $html;
}

ul.pagination li {
    display: inline;
    list-style-type: none;
    padding-right: 1em;
}

<?php echo use_helper('Pagination') ?>
<?php echo pagination($pager) ?>

Comments on this snippet

#1 dav pach on 2006-07-19 at 11:58

Maybe you can make a better CSS to display your pagination but it is nice

#2 Francois Zaninotto on 2006-07-19 at 06:26

I think you could use the link_to_if() helper to display either a link to a page or the sole name of the page without link if it's the current one.

#3 brikou on 2006-07-20 at 11:53

yes francois, i could change the lines (in the foreach loop) by this one

$html .= '<li>' . link_to_unless ($page == $pager->getPage(), $page, $uri . $page) . '</li>';

but we loose the emphasize (done with strong) of the current page...

#4 Pierre Minnieur on 2006-09-02 at 01:49

I found a bug, a stupid bug. The "getCurrentInternalUri()" method won't give a correct route name, if you use action forwarding. I'm using it, because I want to fake modules beeing part of another module.

I use it for a nice navigation - I have the area "account management" where I have the modules groups, roles, actions, apps, modules, users - but I want to know if this module is part of the users-array via $sf_first_module, so I wrote a wrapping action executeForward() in the user module, and the routes contain values like forward_module / forward_ation, so the executeForward action just executes this route forwarding routes.

But if I am on a forwarded page, e.g. I'm listening the roles in the database, and I must paginate them on the end of the list, getCurrentInternalUri() won't give me the forwarded route (user_roles_list), instead I get "forward?page=". That's right, the last action/route was the forwarding, but now I can't paginate.

So, instead, try to use the method gettCurrentRouteName() and add a simple '@' in front of it.

<code>$uri = '@'.sfRouting::getInstance()->getCurrentRouteName();</code>

This simply works.

#5 Pierre Minnieur on 2006-09-02 at 02:03

And I found another problem (omg, it's bug-clean-day!). The route must contain only the page-var, no others, otherwise they won't be put in. So, I added a simple mechanism to add custom route values.

<code>function pagination($pager, $params = array()) { $uri = '@'.sfRouting::getInstance()->getCurrentRouteName(); if (is_array($params) and ($count = count($params)) !== false and $count > 0) { $counter = 0; foreach ($params as $key => $value) { $uri .= (strstr($uri, '?') ? '&' : '?') . $key . '=' . $value; } } ... </code>

That's it. Now you can use it like:

<code><?php echo pagination($users, array('user_role_name' => $list_user_role->getName())) ?></code>

#6 damien pitard on 2009-07-07 at 01:07

with Symfony 1.2 you need to write <pre> $uri = sfContext::getInstance()->getRouting()->getCurrentInternalUri(); instead </pre> instead of <pre> $uri = sfRouting :: getInstance()->getCurrentInternalUri(); </pre>