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我正在尝试更改用户的密码。在下拉列表中选择用户。

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('password', PasswordType::class, array('label' => 'Password'))
        ->add('reset', SubmitType::class, array('label' => 'Reset'))
        ->add('username', EntityType::class, array('class' => User::class,
            'query_builder' => function (EntityRepository $er) {return $er->createQueryBuilder('u')->orderBy('u.username', 'ASC');
            },'choice_label'=>'username', 'label'=>'User'));

}

public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults(array('data_class' => User::class));
}

首次呈现表单时，我会看到用户列表。然后我填写新密码并提交。然后我得到错误：“在属性路径用户名处给出的类型为“字符串”、“对象”的预期参数”。错误发生在控制器中

$form->handleRequest($request);

这是控制器：

    $form = $this->createForm(ResetPwdType::class, new User());

    $form->handleRequest($request);

        if ($form->isSubmitted() && $form->isValid()) {

            $u = $form->get('username')->getData();
            $user = $this->getDoctrine()->getManager()->getRepository(User::class)->findOneBy(['username' => $u->getUsername()]);

            $newPwd = $passwordEncoder->encodePassword($user, $form->get('password')->getData());
            $user->setPassword($newPwd);

            $entityManager = $this->getDoctrine()->getManager();
            $entityManager->persist($user);
            $entityManager->flush();
            $this->addFlash('success', 'Password reset!');
            $params = null;
            return $this->redirectToRoute('reset_pwd');


        }


    return $this->render(
        'security/resetpwd.html.twig',array('form' => $form->createView()));

我认为有一个属性需要是字符串而不是对象。这只是与用户名属性有关。我试图在用户实体上添加一个 toString 方法但没有成功。我必须改变什么才能摆脱错误并将新密码保存到数据库中。

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