Symfony 为表单类型中的数据类赋值
我有一个表单类型,希望知道在我下面的例子中,在Symfony 为表单类型中的数据类赋值,symfony,Symfony,我有一个表单类型,希望知道在我下面的例子中,在setDefaultOptions中针对data\u class应该做什么。我知道我们通常放置实体的路径,但在这种情况下,我嵌入了两个实体,那么我现在该怎么做 我知道我们可以忽略它,但我不想像SensioLabs()建议的那样 表单类型: namespace Car\BrandBundle\Form\Type; use Symfony\Component\Form\AbstractType; use Symfony\Component\Form\F
setDefaultOptions
中针对data\u class
应该做什么。我知道我们通常放置实体的路径,但在这种情况下,我嵌入了两个实体,那么我现在该怎么做
我知道我们可以忽略它,但我不想像SensioLabs()建议的那样
表单类型:
namespace Car\BrandBundle\Form\Type;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;
class BothType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->setMethod('POST')
->setAction($options['action'])
->add('brands', new BrandsType())
->add('cars', new CarsType())
->add('button', 'submit', array('label' => 'Add'))
;
}
public function setDefaultOptions(OptionsResolverInterface $resolver)
{
$resolver->setDefaults(array('data_class' => '?????????????????????'));
}
public function getName()
{
return 'both';
}
}
namespace Car\BrandBundle\Controller;
use Car\BrandBundle\Entity\Brands;
use Car\BrandBundle\Entity\Cars;
use Car\BrandBundle\Form\Type\BothType;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
class BothController extends Controller
{
public function indexAction()
{
$entity = array(new Brands(), new Cars());
$form = $this->createForm(new BothType(), $entity,
array('action' => $this->generateUrl('bothCreate')));
return $this->render('CarBrandBundle:Default:both.html.twig',
array('page' => 'Both', 'form' => $form->createView()));
}
}
控制器:
namespace Car\BrandBundle\Form\Type;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;
class BothType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->setMethod('POST')
->setAction($options['action'])
->add('brands', new BrandsType())
->add('cars', new CarsType())
->add('button', 'submit', array('label' => 'Add'))
;
}
public function setDefaultOptions(OptionsResolverInterface $resolver)
{
$resolver->setDefaults(array('data_class' => '?????????????????????'));
}
public function getName()
{
return 'both';
}
}
namespace Car\BrandBundle\Controller;
use Car\BrandBundle\Entity\Brands;
use Car\BrandBundle\Entity\Cars;
use Car\BrandBundle\Form\Type\BothType;
use Symfony\Bundle\FrameworkBundle\Controller\Controller;
class BothController extends Controller
{
public function indexAction()
{
$entity = array(new Brands(), new Cars());
$form = $this->createForm(new BothType(), $entity,
array('action' => $this->generateUrl('bothCreate')));
return $this->render('CarBrandBundle:Default:both.html.twig',
array('page' => 'Both', 'form' => $form->createView()));
}
}
当我回显提交的数据时,我将获得以下复制数据:
Array
(
[0] => Car\BrandBundle\Entity\Brands Object
(
[id:protected] =>
[name:protected] =>
[origin:protected] =>
)
[1] => Car\BrandBundle\Entity\Cars Object
(
[id:protected] =>
[model:protected] =>
[price:protected] =>
)
[brands] => Car\BrandBundle\Entity\Brands Object
(
[id:protected] =>
[name:protected] => Mercedes
[origin:protected] => Germany
)
[cars] => Car\BrandBundle\Entity\Cars Object
(
[id:protected] =>
[model:protected] => SL500
[price:protected] => 25,000
)
)
我认为最好的方法是忽略它,因为没有特定的实体可以链接它。该文档可能应理解为“如果您的表单绑定到某个实体,则最好是” 在您的情况下,您没有实体类来映射它。所以除非您创建一个,否则应该将其留空。是的,我忽略了控制器中传递的数组,并将
null
分配给data\u class
,这是正常的。