Laravel Lumen 中的自定义 404 页面

发布于02月05日

I'm new to Lumen and want to create an app with this framework. Now I have the problem that if some user enters a wrong url => http://www.example.com/abuot (wrong) => http://www.example.com/about (right), I want to present a custom error page and it would be ideal happen within the middleware level.

Furthermore, I am able to check if the current url is valid or not, but I am not sure how can I "make" the view within the middleware, the response()->view() won't work.

如果有人能帮忙，那就太棒了.

推荐答案

由于错误是App\Exceptions\Handler次处理的，所以这里是处理错误的最佳场所.

如果您只需要一个自定义的404错误页面，那么您可以很容易地做到这一点:

将此行添加到Handler文件的顶部:

use Symfony\Component\HttpKernel\Exception\NotFoundHttpException;

将render函数改为:

public function render($request, Exception $e)
{
    if($e instanceof NotFoundHttpException){
        return response(view("errors.404"), 404);
    }
    return parent::render($request, $e);
}

这假设您的自定义404页面存储在视图中的错误文件夹中，并将返回自定义错误页面和404状态代码.