Question

我正在尝试抛出异常而我正在执行以下操作：

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');

然而我收到的错误就是找不到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");

我认为你有点倒退： - ）

Answer 2

在任何控制器中，您可以将其用于Symfony内的 404 HTTP响应

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

与

相同

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

或 500 HTTP响应代码

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

与

相同

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

或

//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 you can flow this URL

Answer 3

如果它在控制器中，你可以这样做：

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

Answer 4

在控制器中，您只需执行以下操作：

public function someAction()
{
    // ...

    // Tested, and the user does not have permissions
    throw $this->createAccessDeniedException("You don't have access to this page!");

    // or tested and didn't found the product
    throw $this->createNotFoundException('The product does not exist');

    // ...
}

在这种情况下，不需要在顶部包含use Symfony\Component\HttpKernel\Exception\HttpNotFoundException;。原因是你没有直接使用该类，比如使用构造函数。

在控制器外，您必须指明可以找到类的位置，并像往常一样抛出异常。像这样：

use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;

// ...

// Something is missing
throw new HttpNotFoundException('The product does not exist');

或

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

// ...

// Permissions were denied
throw new AccessDeniedException("You don't have access to this page!");

symfony2并抛出异常错误

4 个答案: