我编写了一个函数,该函数具有分别激活的两种形式,并保留同一实体。首先激活输入字段,然后激活随机生成的字段。
但是当我单击“提交”时,它会依次激活两个表单。如果使用条件来防止这种情况出现,但我看来这是行不通的。
我的代码:
$id = $request->get('id');
$user = $this->container->get('account')->getUserRepository()->find($id);
$form1 = $this->createFormBuilder()
->add('password', PasswordType::class, array(
'label' => 'Enter New Password',
'attr' => ['class'=>'form-control']))
->add('save', SubmitType::class, array(
'label' => 'Send', 'attr' => ['class' => 'btn btn-primary action-save']
))
->getForm();
$form2 = $this->createFormBuilder()
->add('password', PasswordType::class, array(
'label' => 'Generate New Password',
'disabled'=> true,
'attr' => ['class'=>'form-control']))
->add('save', SubmitType::class, array(
'label' => 'Send',
'attr' => ['class' => 'btn btn-primary action-save']
))
->getForm();
$form1->handleRequest($request);
if($form1->isSubmitted() && $form1->isValid()) {
$this->addFlash(
'notice',
'You successfully changed the password!'
);
$data = $form1->getData();
$new_password = $data['password'];
$encoder = $this->container->get('security.encoder_factory')->getEncoder($user);
$new_pwd_encoded = $encoder->encodePassword($new_password);
$oneTimePsw = '';
$user->setPassword($new_pwd_encoded);
$manager = $this->getDoctrine()->getManager();
$manager->flush();
}
$form2->handleRequest($request);
if($form2->isSubmitted() && $form2->isValid()) {
$this->addFlash(
'notice',
'Password is successfully generated!'
);
$data = $form2->getData();
$new_password = $data['password'];);
$new = $this->get('member.account')->generateRandomPassword();
$oneTimePsw = '';
$user->setPassword($new);
$manager = $this->getDoctrine()->getManager();
$manager->flush();
}
return $this->render('@AdminTemplates/admin/reset_password.html.twig', array(
'form1' => $form1->createView(),
'form2' => $form2->createView()
));
My twig
<div id="setPassword" style="display:none;">
{{ form_start(form1) }}
{{ form_end(form1) }}
</div>
<div id="generatePassword" style="display:none;">
{{ form_start(form2) }}
{{ form_end(form2) }}
</div>
答案 0 :(得分:0)
我认为您的问题很简单,如果不是其他问题,当我遇到同样的问题时,这就是我所做的事情:
$id = $request->get('id');
$user = $this->container->get('account')->getUserRepository()->find($id);
$form1 = $this->createFormBuilder()
->add('password', PasswordType::class, array(
'label' => 'Enter New Password',
'attr' => ['class'=>'form-control']))
->add('save', SubmitType::class, array(
'label' => 'Send', 'attr' => ['class' => 'btn btn-primary action-save']
))
->getForm();
$form2 = $this->createFormBuilder()
->add('password', PasswordType::class, array(
'label' => 'Generate New Password',
'disabled'=> true,
'attr' => ['class'=>'form-control']))
->add('save', SubmitType::class, array(
'label' => 'Send',
'attr' => ['class' => 'btn btn-primary action-save']
))
->getForm();
$form1->handleRequest($request);
$form2->handleRequest($request);
if($form1->isSubmitted() && $form1->isValid()) {
$this->addFlash(
'notice',
'You successfully changed the password!'
);
$data = $form1->getData();
$new_password = $data['password'];
$encoder = $this->container->get('security.encoder_factory')->getEncoder($user);
$new_pwd_encoded = $encoder->encodePassword($new_password);
$oneTimePsw = '';
$user->setPassword($new_pwd_encoded);
$manager = $this->getDoctrine()->getManager();
$manager->flush();
}else
if($form2->isSubmitted() && $form2->isValid()) {
$this->addFlash(
'notice',
'Password is successfully generated!'
);
$data = $form2->getData();
$new_password = $data['password'];);
$new = $this->get('member.account')->generateRandomPassword();
$oneTimePsw = '';
$user->setPassword($new);
$manager = $this->getDoctrine()->getManager();
$manager->flush();
}
return $this->render('@AdminTemplates/admin/reset_password.html.twig', array(
'form1' => $form1->createView(),
'form2' => $form2->createView()
));
这应该可行,因为用户不能一次提交2个表单。我的isValid()方法也有问题,因此也请尝试进行测试以清除它。