CODEWITHSUNDEEP
javascriptjquerycakephp
NHTorres
NHTorres

Reputation: 1538

form method post and jquery

I'm doing a cakephp form in my view and clicking on the button, I would like to evaluate values before making a certain action, the problem is not me, in another view I have exactly the same code and if it works, but here , do not. However, I do not have any error console.

echo $this->Form->create('Detalle', array('class'=>'form_center', 'type'=>'file','id'=>'detalle_form'));
echo $this->Form->input('FechaEvento', array('type'=>'hidden','value'=>$fecha_evento));
    if($SuperAdmin and !$evento['Evento']['gratuito']){
        echo $this->Form->input('Inscripcion.cortesia', array('type'=>'checkbox', 'value'=>1, 'label' => utf8_encode('Cortesía'), 'div' => 'input checkbox inputRutPasaporte'));
        echo $this->Form->input('Inscripcion.boleta_no_generar', array('type' => 'hidden', 'value' => 0));
    }
$options_btn = array('label' => __('Pagar'), 'class' => 'btn_blue', 'div' => false);
echo $this->Form->end($options_btn);

javascript :

<script type="text/javascript">
$(document).ready(function(){
<? if($evento['Evento']['generar_boleta']){ ?>
    <? if($evento['Evento']['habilitar_compra_multiple']){ ?>   
    $('#detalle_form').data('callback', function(form){
        alert("TEST");
    });
    <? } ?>
<? } ?>
});
</script>

Upvotes: 0

Views: 36

Answers (1)

Karl
Karl

Reputation: 410

The code looks fine but the point is we need information on your control flow. I assume there are missing variables due to it being different. I advise you to

  • combine the code into a CakePHP element that is called in the view so you write the code only once (keeping your code DRY)

  • set DebugLevel to 2

  • install CakePHP DebugKit DebugKit on Github

  • check for missing variables and errors

  • if the problem persists, give us more information accordingly

Upvotes: 1

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