问题描述
如果我想捕获传递给我的网络应用程序的每个单独的非现有网址，并为他们提供不提供404的视图，我该如何做？
If I want to catch every single non existing URL that was passed to my web application and serve them a view without serving a 404, how would I do that?
基本上，我需要记录基于这些命中的统计信息，但我需要通过提供内容，而不是404错误。
Essentially I need to record the statistics based on these hits, but I need to do it by serving content and not a 404 error.
就我可以从 application / config / routes.php 告诉我，我可以使用
As far as I can tell from application/config/routes.php, I could use
$route['default_controller'] = 'catchall';
但我需要的是我的实际网络应用程序。
but I need that for my actual web application.
我也可以使用
$route['404_override'] = 'catchall';
但我不想扔404s。
我尝试使用
$route['(:any)'] = '$1';
但我需要记录整个网址（例如，任何长度的段），而不仅仅是第一段。
but I need to record the entire URL (e.g. any length of segments), not just the first segment.
推荐答案
使用 $ route ['（：any）'] ='catchall_controller'。然后在您的控制器中，您可以使用 $ this-> uri-> segment（n）访问URI段。
Use $route['(:any)'] = 'catchall_controller'. Then in your controller you can access the URI segments using $this->uri->segment(n).
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