Question

我即将提交表单使用Ajax，我已成功使用 POST 提交表单但不知道如何使用Ajax与 Symfony

builform

    $builder->add('name', 'text', array('constraints' => array(new NotBlank()), 'attr' => array('placeholder' => 'Name')))
        ->add('gender', 'choice', array('empty_value' => 'Select Gender', 'constraints' => array(new NotBlank()), 'choices' => \AppBundle\Entity\Records::$gender_list, "required" => true))
        ->add('dateOfBirth', 'birthday', array('label' => 'Date Of Birth','required'=>true))
        ->add('image_path', 'file', array('label' => ' ','required'=>false, 'data_class' => null, 'constraints'=>array(new Assert\File(                                             array('mimeTypes'=>$mime_types, 'maxSize'=>'2048k' )))))
        ->add('country_of_birth', 'entity', array('empty_value' => 'Country of Birth',
            'class' => 'AppBundle\Entity\Location',
            'property' => 'country',
            'label' => 'Country of Birth'
        ))
        ->add('religion', 'entity', array('empty_value' => 'Select Religion',
            'class' => 'AppBundle\Entity\Religion',
            'property' => 'name',
            'label' => 'Religion'
        ));

动作

        $success = false;
        $record_rep = new \AppBundle\Entity\Records();
        $form = $this->createForm(new \AppBundle\Form\AddPersonType(), $record_rep);

        if ($this->getRequest()->getMethod() == 'POST') {
            $form->submit($request);
            if ($form->isValid()) {
                $data = $form->getData();
                $file = $data->getImagePath();
                $image = $file->getClientOriginalName();

                $new_image_name = $this->hanldeUpload($image, $file);
                $this->savetoDB($data, $record_rep, $new_image_name);
                $success = true;
            }
        }
        return $this->render('AppBundle:Homepage:add_person_form.html.twig', array('form' => $form->createView(), 'success'=>$success ));
    }

Answer 1

使用jQuery，使用serialize()表单并将其发布到您的路线。

$('#form').submit(function(e) {

    e.preventDefault();
    var url = "{{ path('YOUR_PATH') }}";
    var formSerialize = $(this).serialize();

    $.post(url, formSerialize, function(response) {
        //your callback here
        alert(response);
    }, 'JSON');
});

在你的行动中

if ($form->isValid()) {

....

  return new Response(json_encode(array('status'=>'success')));
}

一定是这样的。但是你可以在你的路由中添加一些参数，比如格式，方法等。

Answer 2

用于Ajax

 $("#person").submit(function(e){


    var formURL = "{{ path('form') }}";
    var formData = new FormData(this);
    $.ajax({
        url: formURL,
        type: 'POST',
        data:  formData,
        mimeType:"multipart/form-data",
        contentType: false,
        cache: false,
        processData:false,
        success: function(data, textStatus, jqXHR)
        {

        },
        error: function(jqXHR, textStatus, errorThrown)
        {
        }
    });
    e.preventDefault(); //Prevent Default action.
    e.unbind();
});
$("#person").submit();

和For Action

 if ($request->isXmlHttpRequest()) {

....

return new Response(json_encode(array('status'=>'success'));
}

如何在symfony2中提交表单ajax？

2 个答案: