Symfony - 公共和管理部分的不同错误页面

Question

我一直在遵循http://symfony.com/doc/current/cookbook/controller/error_pages.html 上的提示，并在 Resources/TwigBundle/views/Exceptions 中创建了新模板 error500.html.twig。

这很好用，但如果用户位于网站的网络或管理部分，我希望有不同的页面。

有没有简单的方法可以做到这一点？ 谢谢你，迈克。

Answer 1

我认为最好的方法是overriding the default ExceptionController。只需扩展它，并覆盖 findTemplate 方法。从请求的属性中检查是否设置了_route 或_controller，并对其进行处理。

namespace AppBundle\Controller;

use Symfony\Component\HttpFoundation\Request;
use Symfony\Bundle\TwigBundle\Controller\ExceptionController as BaseExceptionController;

class ExceptionController extends BaseExceptionController
{
    protected function findTemplate(Request $request, $format, $code, $showException)
    {
        $routeName = $request->attributes->get('_route');

        // You can inject these routes in the construct of the controller
        // so that you can manage them from the configuration file instead of hardcode them here
        $routesAdminSection = ['admin', 'admin_ban', 'admin_list'];

        // This is a poor implementation with in_array.
        // You can implement more advanced options using regex 
        // so that if you pass "^admin" you can match all the routes that starts with admin.

        // If the route name match, then we want use a different template: admin_error_CODE.FORMAT.twig
        // example: admin_error_404.html.twig
        if (!$showException && in_array($routeName, $routesAdminSection, true)) {
            $template = sprintf('@AppBundle/Exception/admin_error_%s.%s.twig', $code, format);
            if ($this->templateExists($template)) {
                return $template;
            }

            // What you want to do if the template doesn't exist?
            // Just use a generic HTML template: admin_error.html.twig
            $request->setRequestFormat('html');
            return sprintf('@AppBundle/Exception/admin_error.html.twig');
        }

        // Use the default findTemplate method
        return parent::findTemplate($request, $format, $code, $showException);
    }
}

然后配置twig.exception_controller:

# app/config/services.yml
services:
    app.exception_controller:
        class: AppBundle\Controller\ExceptionController
        arguments: ['@twig', '%kernel.debug%']

# app/config/config.yml
twig:
    exception_controller:  app.exception_controller:showAction

然后您可以以相同的方式覆盖模板：

• 资源/AppBundle/视图/异常/
  • admin_error.html.twig
  • admin_error_404.html.twig
  • admin_error_500.html.twig
  • ...

更新

执行此操作的更简单方法是在您的路线的defaults 集合中指定网站部分。示例：

# app/config/routing.yml
home:
    path:      /
    defaults:
        _controller: AppBundle:Main:index
        section:     web
blog:
    path:      /blog/{page}
    defaults:
        _controller: AppBundle:Main:blog
        section:     web
dashboard:
    path:      /admin
    defaults:
        _controller: AppBundle:Admin:dashboard
        section:     admin
stats:
    path:      /admin/stats
    defaults:
        _controller: AppBundle:Admin:stats
        section:     admin

然后你的控制器变成这样：

namespace AppBundle\Controller;

use Symfony\Component\HttpFoundation\Request;
use Symfony\Bundle\TwigBundle\Controller\ExceptionController as BaseExceptionController;

class ExceptionController extends BaseExceptionController
{
    protected function findTemplate(Request $request, $format, $code, $showException)
    {
        $section = $request->attributes->get('section');
        $template = sprintf('@AppBundle/Exception/%s_error_%s.%s.twig', $section, $code, format);
        if ($this->templateExists($template)) {
            return $template;
        }

        return parent::findTemplate($request, $format, $code, $showException);
    }
}

并以与上述相同的方式配置twig.exception_controller。 现在您只需要为每个部分、代码和格式定义一个模板。

• web_error_404.html.twig
• web_error_500.html.twig
• admin_error_404.html.twig
• admin_error_500.html.twig
• 等等……

Answer 2

对于 Symfony 5，这就是我所做的，我相信它也适用于 Symfony 6。它不是很精致，可以改进。

我将vendor/symfony/twig-bridge/ErrorRenderer/TwigErrorRenderer.php 复制到我的应用程序src\CustomerErrorRenderer.php。

有以下区别：

public function __construct(
    Environment       $twig,
    RequestStack      $requestStack,
    HtmlErrorRenderer $fallbackErrorRenderer = null,
                      $debug = false
) {
    if (!\is_bool($debug) && !\is_callable($debug)) {
        throw new \TypeError(
            sprintf(
                'Argument debug passed to "%s()" must be a boolean or a callable, "%s" given.',
                __METHOD__,
                get_debug_type($debug)
            )
        );
    }

    $this->twig = $twig;
    $this->fallbackErrorRenderer = $fallbackErrorRenderer ?? new HtmlErrorRenderer();
    $this->debug = $debug;
    $this->requestStack = $requestStack;
}


private function findTemplate(int $statusCode, RequestStack $requestStack): ?string
{
    $requestUri = $requestStack->getCurrentRequest()
                               ->getRequestUri();

    $prefix = '';
    if ($this->startsWith($requestUri, '/admin/')) {
        $prefix = 'admin/';
    }
    $template = sprintf('@Twig/Exception/%serror%s.html.twig', $prefix, $statusCode);
    if (
        $this->twig->getLoader()
                   ->exists($template)
    ) {
        return $template;
    }

    $template = sprintf('@Twig/Exception/%serror.html.twig', $prefix);
    if (
        $this->twig->getLoader()
                   ->exists($template)
    ) {
        return $template;
    }

    return null;
}


private function startsWith(string $string, string $startString): bool
{
    $len = strlen($startString);
    return (substr($string, 0, $len) === $startString);
}

然后覆盖services.yaml中的error_renderer：

error_renderer:
  class: App\Twig\CustomErrorRenderer
  arguments:
    - '@twig'
    - '@request_stack'
    - '@twig.error_renderer.html.inner'
    - !service
      factory: [ 'Symfony\Component\ErrorHandler\ErrorRenderer\HtmlErrorRenderer', 'getAndCleanOutputBuffer' ]
      arguments: ['@request_stack']

在 routes/framework.yaml、/admin/_error/404、/admin/_error/500 中测试（复制来自 resource: '@FrameworkBundle/Resources/config/routing/errors.xml' 的参数）

when@dev:
_errors:
    path: /_error/{code}.{_format}
    controller: 'error_controller::preview'
    defaults:
        _format: 'html'
    requirements:
        code: '\d+'

_errors_admin:
    path: /admin/_error/{code}.{_format}
    controller: 'error_controller::preview'
    defaults:
        _format: 'html'
    requirements:
        code: '\d+'

还有 config/packages/security.yaml

security:
    firewalls:
        dev:
            pattern: ^/(_(profiler|wdt)|css|images|js|admin\/_error)/
        security: false

在

中创建的模板

templates/bundles/TwigBundle/Exception/admin/
templates/bundles/TwigBundle/Exception/admin/error.html.twig 
templates/bundles/TwigBundle/Exception/admin/error404.html.twig