How can you call a CJuiDialog in beforesave - yii

Question

I will appreciate if somebody could help me to find how to solve out one problem, I have a checkbox in my create form. If i pushed the create button I want to have a popup window if the checkbox is checked and do nothing if the checkbox is unchecked.
my codes in _form
<?php echo $form->checkBoxRow($model, 'default'); ?>
<div class="form-actions">
<?php $this->widget('bootstrap.widgets.TbButton', array(
'buttonType'=>'submit',
'type'=>'primary',
'label'=>$model->isNewRecord ? 'Create' : 'Save',
)); ?>
</div>
in my create controller
public function actionCreate()
{
$model=new EmpSched;
// Uncomment the following line if AJAX validation is needed
// $this->performAjaxValidation($model);
if(isset($_POST['EmpSched']))
{
$model->attributes=$_POST['EmpSched'];
if($model->save())
$this->redirect(array('view','id'=>$model->id_empsched));
}
$this->render('create',array(
'model'=>$model,
'emp'=> new CActiveDataProvider('schedule'),
));
}
in my create.php
<?php
$this->breadcrumbs=array(
'Emp Scheds'=>array('index'),
'Create',
);
$this->menu=array(
array('label'=>'List EmpSched','url'=>array('index')),
array('label'=>'Manage EmpSched','url'=>array('admin')),
);
?>
<h1>Create EmpSched</h1>
<?php echo $this->renderPartial('_form', array('model'=>$model)); ?>
<?php $this->widget('bootstrap.widgets.TbGridView',array(
'id'=>'schedule-grid',
'dataProvider'=>$emp,
'columns'=>array(
'id_sched',
'sched_name',
array(
'name'=>'mon',
'value'=>'$data->fkidday->monday->name'
),
array(
'name'=>'tues',
'value'=>'$data->fkidday->tuesday->name'
),
array(
'name'=>'wed',
'value'=>'$data->fkidday->wednesday->name'
),
array(
'name'=>'thurs',
'value'=>'$data->fkidday->thursday->name'
),
array(
'name'=>'fri',
'value'=>'$data->fkidday->friday->name'
),
/*'sat',
'sun',
*/
),
));
?>
<script>
$(function() {
// when row is clicked
$('#schedule-grid tbody:first').on('click', 'tr', function() {
// get the ID
var id = $(this).find('td:first').text();
// select the same option in dropdown list with same value
$('#EmpSched_fk_id_sched')
.find("option[value="+ id +"]")
.prop("selected", "selected");
});
});
</script>

Answer 1

Add custom javascript (supposing the create button will submit the form)
$('#form-id').submit(function() {
if $(this).find('#check-box-id').is(':checked') {
// open popup here
}
return false;
});