我想要一个关于所有模板的表格。对于显示器,我没有任何担心。我使用一个控制器,并将动作集成到基本模板(通常是base.html.twig)中。下面是控制器的操作:
public function showForm(Request $request)
{
$form = $this->createForm(LetsTalkType::class);
$form->handleRequest($request);
return $this->render('_lets_talk.html.twig', array('form' => $form->createView()));
}下面是与模板的集成:
{{render(controller('App\\Controller\\LetsTalkController::showForm')) }}问题是对于每个控制器(PostController、ServiceController、PortfolioController等)我必须添加一段代码来检查表单是否已发送并有效,然后调用LestsTalkController控制器的操作,如下所示:
public function show($slug, Request $request, LetsTalkController $letsTalkController, MailerInterface $mailer)
{
$letsTalk = new LetsTalk();
$letsTalkForm = $this->createForm(LetsTalkType::class);
$letsTalkForm->handleRequest($request);
if ($letsTalkForm->isSubmitted() && $letsTalkForm->isValid()) {
$letsTalkController->sendEmail($letsTalkForm, $request, $mailer);
}
}没有避免在所有控制器上编写代码的解决方案吗?感谢您的回复:)。
发布于 2021-05-26 01:48:15
更好的方法是将表单的操作设置为将其发送到单个控制器/操作
<?php
class ContactFormController extends AbstractController
{
/**
* @Route("/contact-form", name="contact_form")
*/
public function index(Request $request = null, EmailService $email)
{
$formEntity = new LetsTalk();
$form = $this->createForm(LetsTalkType::class, $formEntity, [
'action' => $this->generateUrl('contact_form'), // set action here to post form to
'method' => 'POST',
]);
if (null !== $request && $request->isMethod('POST')) {
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$formEntity = $form->getData();
...
$email->sendNewContact($formEntity); // or watever
return $this->redirect($request->headers->get('referer')); // redirect to previous page
}
}
return $this->render('_lets_talk.html.twig', [
'form' => $form->createView(),
]);
}
}然后,您可以按照您所做的那样呈现窗体
发布于 2021-05-19 22:03:31
我找到了我的错误。我把它放在下面,因为它可能对处于相同情况的人有用。对于模板中包含的控制器:
{{render(controller('App\\Controller\\LetsTalkController::letsTalkForm', {request: app.request})) }}https://stackoverflow.com/questions/67603333
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