symfony2并抛出异常错误

Question

我试图抛出异常,我正在做以下事情:

use Symfony\Component\HttpKernel\Exception\HttpNotFoundException;
use Symfony\Component\Security\Core\Exception\AccessDeniedException;

然后我按照以下方式使用它们:

 throw new HttpNotFoundException("Page not found");
   throw $this->createNotFoundException('The product does not exist');

但是我收到的错误就像找不到HttpNotFoundException等.

这是抛出异常的最佳方式吗？

Answer 1

尝试:

use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;

和

throw new NotFoundHttpException("Page not found");

我觉得你有点倒退了:-)

Answer 2

在任何控制器中,您都可以在Symfony中使用它来进行404 HTTP响应

throw $this->createNotFoundException('Sorry not existing');

与...一样

throw new NotFoundHttpException('Sorry not existing!');

或者这个500 HTTP响应代码

throw $this->createException('Something went wrong');

与...一样

throw new \Exception('Something went wrong!');

要么

//in your controller
$response = new Response();
$response->setStatusCode(500);
return $response;

或者这是针对任何类型的错误

throw new Symfony\Component\HttpKernel\Exception\HttpException(500, "Some description");

另外...对于自定义例外, 您可以传递此URL

Answer 3

如果它在控制器中,你可以这样做:

throw $this->createNotFoundException('Unable to find entity.');