Reputation: 25
First sorry for my english.
I'm trying to use pagination with ajax with cakePHP 2.8.3.
The first load works good. I see the paginator number :
When I click on page number "2" or on "Next >>", the content #content-recherche is refresh but the link page number 1 (and link "<< Previous") is not usable. The page number "2" stay clickable ...
Here my code in controller :
Public variable for pagination :
public $paginate = array('Entite' => array(
'limit' => 2,
'order' => array(
'Entite.nom' => 'asc'
),
));
The code inside my action:
$this->Paginator->settings = $this->paginate;
$this->Paginator->settings = array(
'conditions' => $conditions,
//'limit' => $sql_limit
'limit' => 2
);
$data = $this->Paginator->paginate('Entite');
$this->set('data', $data);
My code in view:
<div id="content-recherche">
...
</div>
<ul class="pagination">
<?php
$this->Js->JqueryEngine->jQueryObject = 'jQuery';
$this->Paginator->options(array(
'update' => '#content-recherche',
'url' => array('controller' => 'Recherche', 'action' => 'coiffure'),
'complete' => '$.getScript("/js/utils.js", function (data, textStatus, jqxhr) {});',
'evalScripts' => true,
));
?>
<?php
echo $this->Paginator->numbers(array(
'first' => '<<',
'currentClass ' => 'active',
'tag' => 'li',
'modulus' => 5,
'last' => '>>'));
?>
<?php
echo $this->Paginator->prev(
'« Previous', null, null, array('class' => 'disabled')
);
echo $this->Paginator->next(
'Next »', null, null, array('class' => 'disabled')
);
?>
</ul>
Someone see what is my error ?
Thanks.
Upvotes: 0
Views: 141
Reputation: 654
<div id="content-recherche">
...
</div>
<ul class="pagination">
...
</ul>
You are not updating/refreshing code in <ul class="pagination"></ul>
.
You have to refresh/update code in <ul class="pagination"></ul>
tag as well when you refresh/update content in <div id="content-recherche"></div>
block.
Upvotes: 3