Symfony2 发送表单 ajax

Symfony2 send form ajax

提问by Lughino

I'm trying to submit a form via ajax to update a field of an entity, but do not know how to retrieve the data from the controller:

我正在尝试通过 ajax 提交表单以更新实体的字段，但不知道如何从控制器检索数据：

<form class="ajax" action="{{ path('ajax_setSocial') }}" method="post" {{ form_enctype(form) }}>
  <div class="editor">
        {{ form_errors(form) }}
        <div class="editLabel pls">{{ form_label(form.ragSocial) }}</div>
        <div class="editField"> 
            <div class="ptm">
                {{ form_widget(form.ragSocial) }} {{ form_errors(form.ragSocial) }}
            </div>     
            {{ form_rest(form) }}
            <div class="mtm">
                <button class="btn btn-primary disabled save" type="submit">Save</button>
                <button class="btn ann">Close</button>
            </div>
        </div>
  </div>

  var url = Routing.generate('ajax_setSociale');
        var Data = $('form.ajax').serialize();
        $.post(url, 
            Data
            , function(results){
                if(results.success == true) {
                    $(this).parents('ajaxContent').remove();
                    $(this).parents('.openPanel').removeClass('openPanel');
                } else {
                    alert('False'); //test
                }
        });

controller (ajax_setSocial route)

控制器（ajax_setSocial 路由）

public function setSocialeAction(Request $request)
{
      $em = $this->getDoctrine()->getManager();
      // $id = $request->get('form'); ???
    $entity = $em->getRepository('MyBusinessBundle:Anagrafic')->find($id);

    if (!$entity) {
        throw $this->createNotFoundException('Unable to find Anagrafic entity.');
    }

    $form = $this->createFormBuilder($entity)
       ->add('ragSocial', 'text', array('label' => 'Social'))
       ->add('id', 'hidden')
        ->getForm();
    $form->bind($request);

    if ($form->isValid()) {
        $em->persist($entity);
        $em->flush();

        $output = array();
        $response = new Response();
        $output[] = array('success' => true);
        $response->headers->set('Content-Type', 'application/json');
        $response->setContent(json_encode($output));
        return $response;
    }

As recovery values ??and then pass the id to create the query and the other values ??to update the entity? And if the fields do not pass validation, how do I pass the error?

作为恢复值？然后传递id来创建查询和其他值？？更新实体？如果字段未通过验证，我该如何通过错误？

采纳答案by ihsan

I suggest to pass the id to the controller.

我建议将 id 传递给控制器​​。

html:

html：

<form class="ajax" action="{{ path('ajax_setSocial', { 'id': entity.id }) }}" method="post" {{ form_enctype(form) }}>

var url = "{{ path('ajax_setSocial', { 'id': entity.id }) }}";

The controller annotation, parameter, and return value, to get the id:

控制器注解、参数和返回值，获取id：

/**
 *
 * @Route("/{id}", name="ajax_setSocial")
 * @Method("POST")
 */
public function setSocialeAction(Request $request, $id) {
    $em = $this->getDoctrine()->getManager();
    $entity = $em->getRepository('MyBusinessBundle:Anagrafic')->find($id);

    return array(
        'entity' => $entity
    );
}

Passing error back to html is like this:

将错误传回 html 是这样的：

// dummy line to force error:
// $form->get('ragSocial')->addError(new FormError("an error message"));

if ($form->isValid()) {
    ...
} else {
    $errors = $form->get('ragSocial')->getErrors(); // return array of errors
    $output[] = array('error' => $errors[0]->getMessage()); // the first error message
    $response->headers->set('Content-Type', 'application/json');
    $response->setContent(json_encode($output));
    return $response;
}

回答by Lighthart

I think you want this:

我想你想要这个：

symfony2 chained selectors

symfony2 链式选择器

However, this one may also be useful:

但是，这个也可能有用：

Many-to-Many Ajax Forms (Symfony2 Forms)(Answer 3)

多对多 Ajax 表单（Symfony2 表单）（答案 3）