How to show an input after doing $ajax->observeField

I have a form with two inputs:

echo $this->Form->input('articulo_id', array('empty'=>'---Select---'));

echo $this->Form->input('numeral_id', array('empty'=>'---Select---', 'type'=>'hidden'));

When the first one changes state the second one gets updated and auto-populated; this is how it gets updated:

$options = array('url' => 'getArtNum', 'update' => 'EvidenciaNumeralId');

echo $ajax->observeField('EvidenciaArticuloId', $options);

The thing is that I need the second one to be shown after doing the update. Is there a way to do it?

I know if I just delete the 'type'=>'hidden' from the开发者_如何学JAVA form the input will be shown but I need it to stay hidden until the first one change.

Thanks in advance.

---

Maybe it's too late but maybe it helps other people with same problem.

You have to try callbacks:

$options['after'] = "$('EvidenciaNumeralId').show()";
echo $ajax->observeField('EvidenciaArticuloId', $options);

This should help.