Wie man Formular Ajax in symfony2 einreicht?

Question

Ich bin über meine Form mit Ajax einreichen, ich habe meine Form mit POST erfolgreich einreichen aber nicht wissen, wie Ajax zu verwenden, um mit 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' 
     )); 

Aktion

 $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

Mit jQuery, verwenden das Formular und post es auf Ihre Route.

$('#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'); 
}); 

In Ihrer Aktion

if ($form->isValid()) { 

.... 

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

es muss wie dies in Ordnung sein. aber Sie können einige Parameter wie das Format, Methoden usw. in Ihrem Routing hinzufügen.

Answer 2

für die 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(); 

und für Aktion

if ($request->isXmlHttpRequest()) { 

.... 

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