You MUSTdestroy that widget/validator because there is possiblity that
DoctrineForm will hydrate (@see doUpdateObject() method) your object with
VALIDATED hidden input.
2010/12/26 dmitrypol <dmitry...@gmail.com>
> Thank you VERY MUCH. I was not able to make unset work as I think it
> needed the id but you gave me a great idea. I went with this approach
> and using sfWidgetFormInputHidden
>
> if($this->getObject()->isNew())
> {
> $this->setWidgets(array(
> ...
> 'pref_id' => new sfWidgetFormDoctrineChoice(array('model' => $this-
> >getRelatedModelName('pref'), 'add_empty' => false)),
> ...
> }
> else
> {
> $this->setWidgets(array(
> ...
> 'pref_id' => new sfWidgetFormInputHidden(),
> ...
> }
>
> Your comment about security got me thinking. Does anyone have
> experience securing sfWidgetFormInputHidden?
>
> Dmitry
>
>
> On Dec 25, 3:50 pm, Alan Bem <alan....@gmail.com> wrote:
> > If you want to hide any widget, don't outputing them is not a good
> (secure)
> > choice. Here's better idea:
> >
> > In configure() method of your form put
> >
> > if($this->getObject()->isNew()) {
> > $this->widgetSchema['preference_id'] = //etc
> > $this->validatorSchema['preference_id'] = //etc
> >
> > }
> >
> > or in case preference_id is already set
> >
> > if(!$this->getObject()->isNew()) {
> > unset($this['preference_id']);
> >
> > }
> >
> > In templates this should work
> >
> > if(isset($form['preference_id'])) {
> > // output widget
> >
> > }
> >
> > This method is also "echo $form" friendly.
>
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