[英]Symfony3: Construct FormType
I have a form in Symfony3, which I initialize - following the docs - as followed: 我在Symfony3中有一个表单,我按照文档进行了初始化-如下所示:
$form=$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: $ booking是一个已经存在的实体,我想对其进行修改-但我想根据该实体来修改表单-例如:
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: 在先前的Symfony 2.8中,可以像这样构造FormType:
$form=$this->createForm(new BookingType($booking),$booking);
Which is exactly what I want :) But in Symfony3 this method throws an exception. 这正是我想要的:)但是在Symfony3中,此方法引发异常。 How can I pass an entity to my formtype? 如何将实体传递给表单类型?
You can also change a form type based on custom options. 您还可以基于自定义选项更改表单类型。
In the form type: 在表单中输入:
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: 在控制器中:
$form = $this->createForm(BookingType::class, $booking, array(
'localizable' => !$booking->getLocation(),
));
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