Symfony3: Construct FormType

Question

I have a form in Symfony3, which I initialize - following the docs - as followed:

$form=$this->createForm(BookingType::class,$booking);

$booking is an already existing entity, which I want to modify - but I want to modify the form depending on the entity - like:

public function buildForm(FormBuilderInterface $builder,$options) {
    $builder->add('name');
    if(!$this->booking->getLocation()) {
        $builder->add('location');
    }
}

Prior Symfony 2.8 it was possible to construct the FormType like:

$form=$this->createForm(new BookingType($booking),$booking);

Which is exactly what I want :) But in Symfony3 this method throws an exception. How can I pass an entity to my formtype?

You can use form events for this: symfony.com/doc/current/components/form/form_events.html — xabbuh
– xabbuh, Commented Jan 12, 2016 at 9:45
In your formType I think you can do $entity = $builder->getData() — BENARD Patrick
– BENARD Patrick, Commented Jan 12, 2016 at 10:05
@cklm but if you want to change the form configuration based on the entity, you should still use the event system, so you can get strange errors. — Dan Blows
– Dan Blows, Commented Jan 12, 2016 at 13:44

Yassine Guedidi · Accepted Answer · 2016-01-13 09:19:51Z

6

You can also change a form type based on custom options.

In the form type:

public function buildForm(FormBuilderInterface $builder, $options) {
    $builder->add('name');

    if($options['localizable']) {
        $builder->add('location');
    }
}

public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults(array(
        'localizable' => true,
    ));
}

In the controller:

$form = $this->createForm(BookingType::class, $booking, array(
    'localizable' => !$booking->getLocation(),
));

answered Jan 13, 2016 at 9:19

Yassine Guedidi

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1 Comment

Manuel Garcia Over a year ago

Works for me, thanks you so much, I did not remember how to change the default options!

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