如果在 Laravel 5.1 中找不到路由，则显示 404 页面

发布于01月14日

我试图找出显示404页未找到，如果没有找到路由.我听了很多教程，但都不管用.

同样在handler中.php

public function render($request, Exception $e)
{
    if ($e instanceof TokenMismatchException) {
        // redirect to form an example of how i handle mine
        return redirect($request->fullUrl())->with(
            'csrf_error',
            "Opps! Seems you couldn't submit form for a longtime. Please try again"
        );
    }

    /*if ($e instanceof CustomException) {
        return response()->view('errors.404', [], 500);
    }*/

    if ($e instanceof \Symfony\Component\HttpKernel\Exception\NotFoundHttpException)
        return response(view('error.404'), 404);

    return parent::render($request, $e);
}

If I enter wrong URL in browser, it returns a blank page. I have

'debug' => env('APP_DEBUG', true),

in app.php.

如果找不到路由，有人能帮我怎么显示404页面吗？非常感谢.

推荐答案

我收到了500个错误，而不是404个错误.我是这样解决问题的:

在app/Exceptions/Handler.php文件中，有一个render函数.

Replace the function with this function:

public function render($request, Exception $e)
{
    if ($this->isHttpException($e)) {
        switch ($e->getStatusCode()) {

            // not authorized
            case '403':
                return \Response::view('errors.403',array(),403);
                break;

            // not found
            case '404':
                return \Response::view('errors.404',array(),404);
                break;

            // internal error
            case '500':
                return \Response::view('errors.500',array(),500);
                break;

            default:
                return $this->renderHttpException($e);
                break;
        }
    } else {
        return parent::render($request, $e);
    }
}

You can then use views that you save in views/errors/404.blade.php, and so on.