我使用以下库/技术:
JMSSerializer,FOSRestBundle Symfony 3.2 PHP 7.1
当我尝试向POST endpoints 发出POST请求时,我无法使表单生效 .
文件:
Country.php - > POPO实体
CountryType.php
<?php
declare(strict_types = 1);
namespace AppBundle\Form;
use AppBundle\Model\Entity\Country;
use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\Extension\Core\Type\IntegerType;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolver;
class CountryType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder, array $options)
{
$builder->add('name' , TextType::class)
->add('iso_alpha_2_code', TextType::class)
->add('iso_alpha_3_code', TextType::class)
->add('is_numeric_code', IntegerType::class);
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults([
'data_class' => Country::class,
'csrf_protection' => false,
]);
}
/**
* @inheritdoc
*/
public function getName() : string
{
return '';
}
}
CountryController.php postAction
public function postAction(Request $request)
{
$country = new Country();
$form = $this->createForm(CountryType::class, $country);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
die('ok');
// TODO INSERT DATA then redirect
return $this->routeRedirectView('get_country', ['id' => $country->getId()]);
}
return $this->get('fos_rest.view_handler')->handle(View::create($form));
}
问题是它不会输入if块,因为isSubmited()和isValid()方法都返回false . 当我调用 $form->getData()
时它会返回
CountryController.php on line 40: Country {#306 -id: null -name: null -isoAlpha2Code: null -isoAlpha3Code: null -isNumericCode: null }
我提出的要求:
你能告诉我一些错误吗?
2 回答
这是我在使用FOSRest包时提交表单的方式,我希望这会让你朝着正确的方向前进 . 这是一个示例注册操作 . 您会注意到您必须手动提交表单
$form->submit($request->request->all());
然后检查它是否有效我能够使用以下行获取数据