Symfony2-表单类:data_class似乎不起作用

问题描述 投票:0回答:2

我正在使用Form类来帮助生成表单。这是一个密码重置表格。

此表单需要绑定到我的用户实体,一旦提交表单,密码将被更新。

一切都在使用我当前拥有的代码,但是试图剥离一些我认为不需要的代码(由于使用了data_class选项),我的验证失败了(因为表单似乎变得“分离”了)形成实体)

好,所以一些代码:

public function passwordResetAction(Request $request, $key = NULL)
{
    $user = $this->getDoctrine()
            ->getRepository('DemoUserBundle:User\User')
            ->findOneBy(array('resetToken' => $key));

    if (!$user) {
        throw $this->createNotFoundException('No user found for reset token '.$key.'!');
    }

    $form = $this->createForm(new PasswordResetType(), $user);

    if ($request->getMethod() == 'POST') {
        $form->bindRequest($request);

        if ($form->isValid()) {
            $postData = $request->request->get('resetpass');

            $newPassword = $postData['password'];

            $encoder = new MessageDigestPasswordEncoder('sha1', false, 1);
            $password = $encoder->encodePassword($newPassword, $user->getSalt());
            $user->setPassword($password);

            $newToken = base64_encode($user->getUsername().'|'.date('Y-m-d'));
            $user->setResetToken($newToken);

            $em = $this->getDoctrine()->getEntityManager();
            $em->persist($user);
            $em->flush();

            return $this->redirect($this->generateUrl('password_reset_success'));
        }

    }

    return $this->render('DemoUserBundle:User\Reset:password-reset.html.twig', array('key' => $key, 'form' => $form->createView()));
}

我感兴趣的代码是“ $ form = $ this-> createForm(new PasswordResetType(),$ user);”如您所见,我正在使用“表单类型”来构造表单,并向其传递将表单附加到User实体的$ user变量。现在,据我所知,我不需要此参数集,因为我在“表单类型”中设置了实体(请参见下面的代码)

namespace Demo\UserBundle\Form\Type\User;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilder;
use Symfony\Component\Form\FormError;
use Symfony\Component\Form\CallbackValidator;

class PasswordResetType extends AbstractType
{
    public function buildForm(FormBuilder $builder, array $options)
    {
        $builder->add('password', 'password', array('label' => 'New Password'));
        $builder->add('confirmPassword', 'password', array('label' => 'Confirm Password', 'property_path' => false));
        $builder->addValidator(new CallbackValidator(function($form)
            {
                if($form['confirmPassword']->getData() != $form['password']->getData()) {
                    $form['confirmPassword']->addError(new FormError('Passwords must match.'));
            }
        }));
}

public function getDefaultOptions(array $options)
{
    return array(
        'data_class' => 'Demo\UserBundle\Entity\User\User',
    );
}

public function getName()
{
    return 'resetpass';
}

}

因此,此表单应默认附加到用户实体(不需要$ user变量),但不起作用。如果没有传递该变量,则验证就根本行不通(表明它没有被附加到用户实体)

什么给了?大家有什么想法吗?

php symfony
2个回答
0
投票
它并不能使您将对象传递给表单。您认为它将如何知道您要使用哪个对象?

0
投票
public function setDefaultOptions(OptionsResolverInterface $resolver) { $resolver->setDefaults(array( 'data_class' => 'Demo\UserBundle\Entity\User\User', )); }

Symfony 2.7的注意事项:表单API稍有更改,因此您应该改用它:

public function configureOptions(OptionsResolver $resolver) { $resolver->setDefaults(array( 'data_class' => 'Demo\UserBundle\Entity\User\User', )); }

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