CakePHP pagination and the get parameters

Question

I have a problem with the pagination in my search page. When a user search something I have a url like domain.com/search/?s=keyword but paginator gives me links lik

Answer 1

I know this is old but found a simple solution that works for me. Add following in view file -

$paginator->options(array('url' => array_merge($this->passedArgs,
array('?' => ltrim(strstr($_SERVER['QUERY_STRING'], '&'), '&')))));

Found it here

Answer 2

I used Matyas solution, but for $keyword I did like this:

$queryString = explode('?', $_SERVER['REQUEST_URI']);
$options = array('url'=> array('controller' => 'post', 'action' => 'search', '?' => $queryString[1]));

Answer 3

To pass all URL arguments to paginator functions, add the following to your view: Plain Text View

$paginator->options(array('url' => $this->passedArgs));

That's it. See http://book.cakephp.org/2.0/en/core-libraries/components/pagination.html

Answer 4

create the options array

$options = array(
    'url'=> array(
        'controller' => 'posts', 
        'action' => 'search', 
        '?' => 'keyword='.$keyword
     )
);

set it to the helper

$paginator->options($options)

and then you can use the paginator helper while retaining the GET variables.

hope that it helped :)

to make it easier you can putl $paginator options in your view or .ctp file

$this->Paginator->options['url']['?'] = $this->params['ur];

then put the value that you want :)

Answer 5

$this->Paginator->options['url']['?'] = ['key' => $this->params['url']['value']];

Answer 6

Basically you can do it like this:

function list_results($keywords)
  {
  $data = $this->paginate('Posts', array('Post.title LIKE' => '%'.$keywords.'%'));
  }