Never create a new response object
there is already one available in your controller
just use
$this->response->body($content);
as documented
Am Freitag, 16. August 2013 19:16:32 UTC+2 schrieb cesar calvo:
>
> I use this in my AppController
>
> public function jsonResponse($array) {
> return new CakeResponse(array('body' => json_encode($array)));
> }
>
> Then on a controller call jsonResponse
>
>
> Note: if you are usin Security component on beforeFilter:
>
> if ($this->request->is('ajax')) $this->Security->unlockedActions =
> array($this->request->action);
>
> On Thursday, August 15, 2013 11:03:34 PM UTC-3, Renato Bigliazzi wrote:
>>
>> Hi , I can not do the twitter bootstrap component typeahead work with
>> cake. i use https://github.com/rudylee/cbunny , but dont work form me.
>>
>>
>> In my view
>>
>> JS
>>
>> <script type="text/javascript">
>> $(document).ready(function(){
>> $('#itemdesc').typeahead({
>> source: function (query, process) {
>> return $.ajax({
>> url:'<?php echo
>> Router::url(array('controller'=>'Invoices','action'=>'localizaprodutos'));?>',
>> type: 'get',
>> data: {q: query},
>> dataType: 'json',
>> success: function (json) {
>> return process(json);
>> }
>> });
>> }
>> });
>> });
>> </script>
>>
>> HTML
>> <input type="text" name="itemdesc[]" class="input-large" id="itemdesc"
>> data-provide="typeahead"/>
>>
>>
>> and controller
>>
>> public function localizaprodutos(){
>> $this->autoRender = false;
>> $this->RequestHandler->respondAs('json');
>>
>> // get the search term from URL
>> $term = $this->request->query['q'];
>> $users =
>> $this->Invoice->Invoicedetail->Inventoryitem->find('all',array(
>> 'conditions' => array(
>> 'Inventoryitem.desc LIKE' => '%'.$term.'%'
>> )
>> ));
>>
>> // Format the result for select2
>> $result = array();
>> foreach($produtos as $key => $produto) {
>> array_push($result, $produto['Inventoryitem']['desc']);
>> }
>> $produtos = $result;
>>
>> echo json_encode($produtos);
>> }
>>
>>
>> Thanks
>>
>> Renato
>>
>>
>>
>>
>>
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