Symfony 嵌入式表单和约束

Question

想象一个像这样的简单实体

class User {
  /**
   * @ORM\Embedded(class="Role")
   */
  protected $role;
}

和这样的表格类型

class UserType extends AbstractType
{
  /**
   * @param FormBuilderInterface $builder
   * @param array $options
   */
  public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder->add('role', RoleType::class);
}

/**
 * @param OptionsResolver $resolver
 */
public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults(
        array(
            'data_class' => 'XXXBundle\Entity\User',
            'csrf_protection' => false,
        )
    );
}
}

和 Role 实体的 RoleType 形式

/**
 * @ORM\Embeddable()
 */
class Role
{
    /**
     * @ORM\Column(type="string")
     * @Assert\Choice(choices={"foo", "bar"})
     */
    private $name;

    /**
     * Role constructor.
     * @param string $name
     */
    public function __construct($name)
    {
        $this->name = $name;
    }

    /**
     * @return string
     */
    public function getName(): string
    {
        return $this->name;
    }
}

class RoleType extends AbstractType
{
    /**
     * @param FormBuilderInterface $builder
     * @param array $options
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder->add('name');
    }

    /**
     * @param OptionsResolver $resolver
     */
    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults(
            array(
                'data_class' => 'XXXBundle\Entity\Role',
                'empty_data' => function (FormInterface $form) {
                    return new Role($form->get('name')->getData());
                },
            )
        );
    }
}

使用典型的形式->isValid，选择约束断言无法正常工作...实际上并未验证角色名称值。

Answer 1

我找到了 Symfony3 的解决方案

    /**
     * @ORM\Embedded(class="Role")
     * @Assert\Valid()
     */
    protected $role;

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