我的后端有一个简单的用户管理器，我希望能够在不设置新密码/每次都重复旧密码的情况下编辑用户。
现在，如果我在编辑用户时将密码字段留空，symfony2 会抱怨必须输入密码，当然我在注册新用户时想要这个功能，但是当我编辑它们时，我想要表单如果未填写密码框，请忽略它们。
这是如何实现的？
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15864 次
8 回答
27
为了其他人的参考，我以这种方式解决了这个问题。
我的表单类型：
public function buildForm(FormBuilder $builder, array $options)
{
$builder
->add('username', 'text', array('label' => 'Servernamn '))
->add('plainPassword', 'repeated', array('type' => 'password', 'first_name' => 'Lösenord för server ', 'second_name' => 'Upprepa lösenord för server',));
$builder-> addValidator(new CallbackValidator(function(FormInterface $form){
$username = $form->get('username')->getData();
if (empty($username)) {
$form['username']->addError(new FormError("Du måste ange ett namn för servern"));
}
}));
}
我的更新操作：
public function updateServerAction($id)
{
$em = $this->getDoctrine()->getEntityManager();
$entity = $em->getRepository('BizTVUserBundle:User')->find($id);
if (!$entity) {
throw $this->createNotFoundException('Unable to find Container entity.');
}
$originalPassword = $entity->getPassword();
$editForm = $this->createForm(new editServerType(), $entity);
$request = $this->getRequest();
$editForm->bindRequest($request);
if ($editForm->isValid()) {
$plainPassword = $editForm->get('plainPassword')->getData();
if (!empty($plainPassword)) {
//encode the password
$encoder = $this->container->get('security.encoder_factory')->getEncoder($entity); //get encoder for hashing pwd later
$tempPassword = $encoder->encodePassword($entity->getPassword(), $entity->getSalt());
$entity->setPassword($tempPassword);
}
else {
$entity->setPassword($originalPassword);
}
$em->persist($entity);
$em->flush();
return $this->redirect($this->generateUrl('Server'));
}
因此，如果设置了我的用户密码，则更新我的用户密码，否则保留原始密码。
于 2012-08-17T21:08:01.507 回答
16
在您的密码属性的实体设置器上，使其如下所示：
/**
* Set password
*
* @param string $password
* @return User
*/
public function setPassword($password)
{
if (!is_null($password)) {
$this->password = $password;
}
return $this;
}
诀窍是检查传递的参数是否为空，如果不是则仅设置。
于 2014-08-29T14:59:36.950 回答
4
如果您需要将所需选项添加到 $options 数组。例子：
public function buildForm(FormBuilder $builder, array $options)
{
$builder
->add('username', 'text', array('label' => 'Servernamn '))
->add('plainPassword', 'repeated', array(
'type' => 'password',
'first_name' => 'password',
'second_name' => 'repeat_password',
'required' => false,
));
}
于 2012-12-20T14:45:04.880 回答
2
我所做的是保持UserType原样，只删除控制器中的密码字段：
/**
* @Route("/users/edit/{username}", name="user_edit")
*/
public function editAction(Request $request, User $user)
{
$form = $this->createForm(UserType::class, $user);
// Remove the password field when editing the user
// Otherwise it must be entered each time the user is edited
// We can change the password via a special edit-user-password page
$form->remove('password');
$form->handleRequest($request);
if ($form->isValid()) {
// ...
}
}
这样你就可以有一个单一的 UserType，在 editAction 和 newAction 中可以重用。
请务必在 .$form->remove('password');之前添加该行$form->handleRequest($request);。
于 2016-07-15T11:12:24.313 回答
0
有一篇文章提供了许多选项来满足您的要求，来自 Symfony2 食谱，特别是以下部分对我有用：
下面是实现：
public function buildForm(FormBuilderInterface $builder, array $options)
{
// ...
$builder->addEventListener(FormEvents::PRE_SET_DATA, function(FormEvent $event) {
$product = $event->getData();
$form = $event->getForm();
// check if the Product object is "new"
// If no data is passed to the form, the data is "null".
// This should be considered a new "Product"
if (!$product || null === $product->getId()) {
$form->add('name', 'text');
}
});
}
于 2014-03-23T08:30:21.783 回答
0
Symfony 5 方法：
在控制器类中：
/**
* @Route("/user/{id}/edit")
*/
public function updateUser(
$id, // See 'Route' annot., needed to `find()` User, adjust to your needs
User $user,
Request $request,
UserPasswordEncoderInterface $passwordEncoder,
UserRepository $userRepository
): Response {
$form = $this->createForm(UserFormType::class, $user);
// if password field is required by default, we have to `add()` it again
// with `required` set to `false`
$form->add(
'plainPassword',
PasswordType::class,
[
// do not use Constraint NotBlank()
'required' => false,
'mapped' => false
]
);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$entity = $userRepository->find($id); // as above, adjust to your needs
if (!$entity) {
throw $this->createNotFoundException('User does not exist!');
}
$originalPassword = $entity->getPassword();
/** @var User $user */
$user = $form->getData();
// keep original password if new one isn't provided
if (!empty($form['plainPassword']->getData())) {
$user->setPassword(
$passwordEncoder->encodePassword(
$user,
$form['plainPassword']->getData()
)
);
} else {
// set old password
$user->setPassword($originalPassword);
}
$em = $this->getDoctrine()->getManager();
$em->persist($user);
$em->flush();
return new Response('User updated!');
}
return $this->render(
'admin/user_edit.html.twig',
[
'requestForm' => $form->createView(),
]
);
}
于 2020-09-08T22:06:46.963 回答
-1
对于验证，使用 2 个不同的验证组并通过可调用函数设置正确的一个。
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add(
'oldPassword',
PasswordType::class,
[
'constraints' => new UserPassword([
'groups' => 'profile_password',
]),
'mapped' => false,
'required' => false,
]
);
$builder->add(
'plainPassword',
PasswordType::class,
[
'constraints' => new NotBlank([
'groups' => 'profile_password',
]),
'mapped' => false,
'required' => false,
]
);
}
/**
* {@inheritdoc}
*/
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults(
[
'validation_groups' => function (FormInterface $form) {
$newPassword = $form->get('plainPassword')->getData();
$oldPassword = $form->get('oldPassword')->getData();
if ($oldPassword || $newPassword) {
return ['profile', 'profile_password'];
} else {
return ['profile'];
}
},
]
);
}
于 2016-09-05T09:23:27.717 回答
-4
你可以用
if (empty($password)) {
//edit the password
}
尝试将其修复到表单的 PHP 中。
于 2012-08-14T21:57:27.713 回答