我通过AJAX(jQuery)发送了一个Symfony表单。 然而,即使数据被明确地发送,Symfony称该形式为“未提交”(实际上没有错误,但是一些调试显示isSubmitted返回false)。使用AJAX调用发送数据后,Symfony表单不会被提交?
这是在我的控制器代码
public function emailrequestFormAction(Request $request)
{
$result = new Result();
$view = new View\EmailRequestView();
$form = $this->createForm(new Type\formType(), $emailrequestview, array(
'action' => $this->generateUrl('ajax_form_emailrequest'),
'method' => 'POST'
));
return $this->render('AppBundle:Account/Form:formemailrequest.html.twig', array('form' => $form->createView()));
}
这是我formType
class formType extends AbstractType {
public function getName()
{
return 'formType';
}
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder
->add('Name', 'text')
->add('Email', 'email')
->add('SendRequest', 'submit', array('label' => 'Invite me'))
->getForm();
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults(array(
'data_class' => 'Bundle\AppBundle\Form\View\EmailRequestView',
'csrf_protection' => false,
));
}}
的代码,这是我的EmailRequestView.php
class EmailRequestSignUpView {
/**
* @Assert\NotBlank(message = "Name should not be blank")
*/
protected $Name;
/**
* @Assert\NotBlank(message = "Email should not be blank")
*/
protected $Email;
/**
* @return mixed
*/
public function getName(){ return $this->Name; }
/**
* @param mixed $Name
*/
public function setName(String $Name){ $this->Name = $Name; }
/**
* @return mixed
*/
public function getEmail(){ return $this->Email; }
/**
* @param mixed $Email
*/
public function setEmail(String $Email){ $this->Email = $Email; } }
现在使用Ajax调用函数我插在我的HTML网页代码提交一个电话,这是JS代码:
function formSubmitAjaxForm(form, e) {
e.preventDefault();
var $this = $(form);
var formType = $this.attr("method");
var formUrl = $this.attr("action");
var formData = $this.serialize();
$.ajax({type: formType, url: formUrl, data: formData})
.done(function (data) {
var Result = data;
try {
// some code
}
catch (err) {
// some code
}
finally {
// some code
}
})
.fail(function (jqXHR, textStatus, errorThrown) {
// some code
}); }
AJAX调用我的控制器ajax_emailrequestFormAction功能。
public function ajax_emailrequestFormAction(Request $request)
{
$result = new Result();
$emailrequestview = new View\EmailRequestView();
$form = $this->createForm(new Type\EmailRequestType(), $emailrequestview, array('method' => 'POST'));
$form->handleRequest($request);
$data = $form->getData();
/* Check if Method is POST */
if ($request->isMethod('POST') == false) {
$result->setSuccessful(false);
$result->setMessage('You can access use Post Method!');
return $result->ResponseJsonResult(400);
}
/* Check if Form is Submitted */
if ($form->isSubmitted() === false) {
$result->setSuccessful(false);
$result->setMessage('Form not submitted.');
return $result->ResponseJsonResult(200);
}
/* Check if Form is Valid */
if ($form->isValid() === false) {
$result->setSuccessful($form->isValid());
$result->setMessage('Form is not valid.');
return $result->ResponseJsonResult(200);
}
/* DataBase Operation */
$result->setContext(', Name: ' . $form->get('Name')->getData() . ', Email: ' . $form->get('Email')->getData());
$result->setMessage('Request send Successfully.');
return $result->ResponseJsonResult(200);
}
* 结果是:成功:假和消息:表单是不是valid.Details:ERROR:名称不能为空。错误:电子邮件不应该是空白 *
现在我试着编写这样的功能:
$form->submit($request);
$form->handleRequest($request);
$form->submit($request->request->get($form->getName()));
,但我没有找到的正确方法得到的只是提交了形式的价值。
有人可以帮我吗? (注:Symfony的版本是2.7)