php - 在 twig symfony 中显示所有表单错误

Question

我有一个包含集合类型(字段 col)的表单。

在 collectiontype 的字段上(称之为 a)，我有返回错误的验证器。

我尝试使用 {{form_errors(form)}} 和 {{form_errors(form.col)}} 但它们不起作用...

Answer 1

你可以使用FormErrorsSerializer获取所有错误并将其作为 JSON 数组返回，此代码片段来 self 正在处理的项目

function addOwnership(Request $request)
{
    $id = $request->get('land_id');
    $land = $this->getDoctrine()->getRepository('DamanBundleCoreBundle:Land')->find($id);
    $ownership = new OwnershipHistory();
    $ownership_form = $this->createForm('Daman\Bundle\CoreBundle\Form\OwnershipType', $ownership);
    $ownership_form->handleRequest($request);

    if ($ownership_form->isSubmitted()) {
        if ($ownership_form->isValid()) {

            $em = $this->getDoctrine()->getManager();
            $ownership->setLandId($land);
            $em->persist($ownership);
            $em->flush();
            $response = array('success' => true,
                'msg' => $this->get('translator')->trans('data_has_been_successfully_added'),
                'id' => $ownership->getId(),
                'action' => 'refresh'
            );

        } else {
            $errors = $this->get('form_serializer')->serializeFormErrors($ownership_form, true, false);
            $response = array('success' => false, 'msg' => $this->get('translator')->trans('error_exist_in_the_form'), 'errors' => $errors);

        }
        $jsonResponse = new JsonResponse($response);
        return $jsonResponse;

    }


}//end function