Symfony 3 - 如何使用表单处理JSON请求

I'm having a hard time figuring out how to handle a JSON request with Symfony forms (using v3.0.1).

我很难弄清楚如何使用Symfony表单处理JSON请求(使用v3.0.1)。

Here is my controller:

这是我的控制器:

/**
 * @Route("/tablet")
 * @Method("POST")
 */
public function tabletAction(Request $request)
{
    $tablet = new Tablet();
    $form = $this->createForm(ApiTabletType::class, $tablet);
    $form->handleRequest($request);

    if ($form->isValid()) {
        $em = $this->getDoctrine()->getManager();
        $em->persist($tablet);
        $em->flush();
    }

    return new Response('');
}

And my form:

我的表格:

class ApiTabletType extends AbstractType
{
    public function buildForm(FormBuilderInterface $builder, array $options)
    {
        $builder
            ->add('macAddress')
        ;
    }

    public function configureOptions(OptionsResolver $resolver)
    {
        $resolver->setDefaults([
            'data_class' => 'AppBundle\Entity\Tablet'
        ]);
    }
}

When I send a POST request with the Content-Type header properly set to application/json, my form is invalid... all fields are null.

当我发送POST请求并将Content-Type标头正确设置为application / json时,我的表单无效...所有字段都为空。

Here is the exception message I get if I comment the if ($form->isValid()) line :

如果我评论if($ form-> isValid())行,这是我得到的异常消息:

An exception occurred while executing 'INSERT INTO tablet (mac_address, site_id) VALUES (?, ?)' with params [null, null]:

使用params [null,null]执行'INSERT INTO tablet(mac_address,site_id)VALUES(?,?)'时发生异常:

I've tried sending different JSON with the same result each time:

我每次尝试发送不同的JSON,结果相同:

{"id":"9","macAddress":"5E:FF:56:A2:AF:15"}
{"api_tablet":{"id":"9","macAddress":"5E:FF:56:A2:AF:15"}}

"api_tablet" being what getBlockPrefix returns (Symfony 3 equivalent to form types getName method in Symfony 2).

“api_tablet”是getBlockPrefix返回的内容(Symfony 3相当于Symfony 2中的表单类型getName方法)。

Can anyone tell me what I've been doing wrong?

谁能告诉我我做错了什么?

UPDATE:

I tried overriding getBlockPrefix in my form type. The form fields have no prefix anymore, but still no luck :/

我尝试在表单类型中覆盖getBlockPrefix。表单字段不再有前缀,但仍然没有运气:/

public function getBlockPrefix()
{
    return '';
}

3 个解决方案

#1

$data = json_decode($request->getContent(), true);
$form->submit($data);

if ($form->isValid()) {
    // and so on…
}

#2

I would recommend you use the FOSRestBundle.

我建议你使用FOSRestBundle。

Here's an example config I use at the moment:

这是我目前使用的示例配置:

fos_rest:
    view:
        view_response_listener: force
        force_redirects:
          html: true
        formats:
            jsonp: true
            json: true
            xml: false
            rss: false
        templating_formats:
            html: true
        jsonp_handler: ~
    body_listener: true
    param_fetcher_listener: force
    allowed_methods_listener: true
    access_denied_listener:
        json: true
    format_listener:
        enabled: true
        rules:
            - { path: ^/, priorities: [ 'html', 'json' ], fallback_format: html, prefer_extension: true }
    routing_loader:
            default_format: ~
            include_format: true
    exception:
        enabled: true
        codes:
            'Symfony\Component\Routing\Exception\ResourceNotFoundException': 404
            'Doctrine\ORM\OptimisticLockException': HTTP_CONFLICT
        messages:
            'Symfony\Component\Routing\Exception\ResourceNotFoundException': true

Make sure you have getBlockPrefix defined with something (I haven't tried empty strings so it might work):

确保你已经定义了getBlockPrefix(我没有尝试过空字符串,所以它可能有效):

public function getBlockPrefix()
{
   return 'api_tablet'
}

Disable CSRF protection for good measure:

禁用CSRF保护措施:

public function configureOptions(OptionsResolver $resolver)
{
    $resolver->setDefaults([
        'csrf_protection' => false,
        'data_class' => 'AppBundle\Entity\Tablet'
    ]);
}

You can then POST the following data to the form:

然后,您可以将以下数据发布到表单:

{"api_tablet":{"macAddress":"5E:FF:56:A2:AF:15"}}

Notes / caveats:

备注/警告:

You need to make sure your Form is configured with all the fields of the Entity. You can configure what you need in a validator.

您需要确保您的表单配置了实体的所有字段。您可以在验证器中配置所需的内容。
You have to post in camelCase. FOSRestBundle supports Array Normalization, which I haven't tried, but apparently that will let you post in underscores.

你必须在camelCase中发帖。 FOSRestBundle支持阵列规范化,我还没有尝试过,但显然会让你发布下划线。

#3

I guess you can drop forms and populate->validate entities

我想你可以删除表单并填充 - >验证实体

$jsonData = $request->getContent();
$serializer->deserialize($jsonData, Entity::class, 'json', [
    'object_to_populate' => $entity,
]);
$violations = $validator->validate($entity);

if (!$violations->count()) {

    return Response('Valid entity');
}

return new Response('Invalid entity');

https://symfony.com/doc/current/components/serializer.html#deserializing-in-an-existing-object

https://symfony.com/components/Validator

#1

$data = json_decode($request->getContent(), true);
$form->submit($data);

if ($form->isValid()) {
    // and so on…
}

#2