Re: CakePHP 2 + JQuery Easy UI

hi, chopite, tries with this code:
 

 function getPaises(){

   

       $paises = $this->Pais->find(

           'all',

           array(

               'fields'=>array(

                   'Pais.id',

                   'Pais.nombre'

               )

           )

       );

$this->set('return',$paises);

$this->render('getPaises','ajax');

  }

  

  

in /View/{Controller}/get_paises.ctp , only this code:

<?php 

if(isset($return)){

echo json_encode($return);

} 

?>

2012/4/28 wchopite <wchopite@gmail.com>

Hi people, I´m from Venezuela sorry my english.

I have a little problem and I need your help

I need to work with Jquery Easy UI (http://www.jeasyui.com) and with
Cakephp together.

But I can´t do it work.

I make the query in the controller and return the data in Json format:

 function getPaises(){
       $this->autoRender= false;
       Configure::write('debug', 0);

       $paises = $this->Pais->find(
           'all',
           array(
               'fields'=>array(
                   'Pais.id',
                   'Pais.nombre'
               )
           )
       );

       return (json_encode($paises));
   }

In view, I have:

<table id="tt"></table>


<input type="hidden" id="url-listado-paises" value="<?php echo $this-
>Html->url(array('controller'=>'paises','action'=>'getPaises')); ?>"/>

<script>
   $('#tt').datagrid({
       url:$('#url-listado-paises').val(),
       columns:[[
           {
               'Pais': {
                   field:'id',title:'ID',width:100,
                   field:'nombre','title':'Nombre'
               }
           }
       ]]
   });
</script>

But it doesn´t work

Any idea why?

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