Question

In my view I am using this code:

$numbers = $this->Paginator->numbers(array(
    'separator'     => '',
    'tag'           => 'li',
    'currentClass'  => 'active'
));

which outputs:

<li class="active">1</li>
<li><a href="/controller/action/page:2">2</a></li>
<li><a href="/controller/action/page:3">3</a></li>

This works pretty well, the only issue I have with it is that the current page is not a link. Is it possible to make the current page a link?

Thanks for reading.

Answer 1

I implemented a class extension like Dave suggested, but instead of copying all the code from the original class, I did a string replace method instead, that way, if I update the CakePHP core library, this should fail pretty gracefully, where as copying all the code from the original Helper may cause loss of features, bug fixes, etc. Here is the class that I implemented:

<?php

class AppPaginatorHelper extends PaginatorHelper
{
    public function numbers($options = array()) {
        $output = parent::numbers($options);

        // get the current page number, and create a link with it
        $current = $this->current();
        $currentLink = $this->link($current, array('page' => $current));

        // if you're using cake pre 2.1 you cannot change the current class with
        // the options array, so it will always be "current"
        $find = "<li class=\"current\">{$current}</li>";
        $replace = "<li class=\"active\">{$currentLink}</li>"; 

        $output = str_replace($find, $replace, $output);

        return $output;
    }
}