Symfony2 MongoDB中的表单

Question

我有一个成员类：

class Member {
/**
 * @ODMId
 */
protected $id;
/**
 * @ODMString
 */
protected $fName;
/**
 * @ODMString
 */
protected $lName;
/**
 * @ODMString
 */
protected $email;

/**
 * @ODMString
 */
protected $password;
//getters & setters
public function __toString()
{
    return $this->getFName().$this->getLName();
}

这个FormType的形式是：

use SymfonyComponentFormAbstractType;
use SymfonyComponentFormFormBuilderInterface;
use SymfonyComponentOptionsResolverOptionsResolverInterface;
class MemberType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('fName', 'text')
            ->add('lName', 'text')
            ->add('password', 'password')
            ->add('email', 'email')
            ->add('sigIn', 'submit');
        $builder->add('terms', 'checkbox', array('property_path' => 'termsAccepted'));
    }
    public function setDefaultOptions(OptionsResolverInterface $resolver)
    {
        $resolver->setDefaults(array(
            'data_class' => 'AcmeMainBundleDocumentMember',
        ));
    }
    public function getName()
    {
        return 'member';
    }
}

在控制器中，我使用它来呈现形式：

public function indexAction()
{
    $form = $this->createForm(new MemberType(), 'member');
    return $this->render('AcmeMainBundle:Default:index2.html.twig',
        array('register'=>$form->createView()));
}

它看起来可以工作，但当我检查它时，我在浏览器中看到了这个错误：

The form's view data is expected to be an instance of class
PrexterMainBundleDocumentMember, but is a(n) string. You can avoid this error
by setting the "data_class" option to null or by adding a view transformer that
transforms a(n) string to an instance of PrexterMainBundleDocumentMember.

我已经将data_class设置为null，但没有任何更改！请帮忙。。。

Answer 1

您应该传递Member的实例作为Controller#createForm()的第二个参数，而不是字符串"member"。例如：

$form = $this->createForm(new MemberType(), new Member());