Thanks,

this is what I was trying to do, but actually it is not very nice :).
It didn't occur to me I could pass a parameter in the constructor.

Paolo

On Sep 8, 4:43 pm, Christopher Schnell <[email protected]>
wrote:
> hi,
>
> you could do something nasty like
>
> if(!$this->getUser()->hasAttribute('user'))
>         $tempSchema=$this->form->getWidgetSchema();
>       unset($tempSchema ['mine'];
>       $this->form->setWidgetSchema($tempSchema);
>
> }
>
> but this wouldn't be testable.
>
> better solution 
> ishttp://eatmymonkeydust.com/2009/08/symfony-forms-flexible-widgets-bas...
>
> Regards,
> Christopher.
>
> > -----Ursprüngliche Nachricht-----
> > Von: [email protected] [mailto:symfony-
> > [email protected]] Im Auftrag von torok84
> > Gesendet: Mittwoch, 8. September 2010 16:32
> > An: symfony users
> > Betreff: [symfony-users] Removing a field from a form in an action
>
> > Hi,
>
> > I have a form to select some search options. I have a field that I
> > wanto to show only if the user is logged in. I want to do something
> > like this:
>
> > if(!$this->getUser()->hasAttribute('user'))
> >       unset($this->form->widgetSchema['mine']);
>
> > that doesn't work because widgetSchema is protected. Moreover doing
> > this would leave a dangling validator for the 'mine' field. I
> > undestand I can create a second form class without the 'mine' field,
> > but it would be nice if I could simply unset a field.
>
> > Thanks
> > Paolo
>
> > --
> > If you want to report a vulnerability issue on symfony, please send it
> > to security at symfony-project.com
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