Problem with 404 code status and custom 404 page

nepster-web · July 18, 2020, 7:29pm

So. I want create custom 404 page with my template.
In my template I use twig function path().

mezzio/mezzio-twigrenderer/blob/c938530f54728fe3102df18079fa0434e627692e/src/TwigExtension.php#L78


}

/**
 * @return TwigFunction[]
 */
public function getFunctions() : array
{
    return [
        new TwigFunction('absolute_url', [$this, 'renderUrlFromPath']),
        new TwigFunction('asset', [$this, 'renderAssetUrl']),
        new TwigFunction('path', [$this, 'renderUri']),
        new TwigFunction('url', [$this, 'renderUrl']),
    ];
}

/**
 * Render relative uri for a given named route
 *
 * Usage: {{ path('article_show', {'id': '3'}) }}
 * Generates: /article/3
 *

If I pass to my 404 page I get exception

github.com

hugocfranciscon/atividade-zend-expressive/blob/416ec668bff98f859a98a552878e535fd82efbfb/vendor/zendframework/zend-expressive-helpers/src/UrlHelper.php#L153



/**
 * @param array $params
 * @param RouteResult $result
 * @return string
 * @throws RenderingException if current result is a routing failure.
 */
private function generateUriFromResult(array $params, RouteResult $result)
{
    if ($result->isFailure()) {
        throw new Exception\RuntimeException(
            'Attempting to use matched result when routing failed; aborting'
        );
    }

    $name   = $result->getMatchedRouteName();
    $params = array_merge($result->getMatchedParams(), $params);
    return $this->router->generateUri($name, $params);
}

/**

What is this behavior for?

it prevents me from using a nice 404 template with links.

nepster-web · July 18, 2020, 7:42pm

I think in this case it would be good practice to use the current url.

Basically, this will work (for generateUriFromResult):

...
    if ($result->isFailure()) {
       return $_SERVER['PATH_INFO'] ?? '/';
    }
...

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Problem with 404 code status and custom 404 page

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