Question

我使用密码调子。

我想做的是,

如果访问网页不存在

例如:com/idontexist

然后,我想首先核对一个数据库,看idontexist是否在数据库中。

如果是这样,我就想把用户带往用户。

例如:com/action/idontexist。

我如何能够这样做?

Answer 1

One solution would be to extend the CI_Router class with your own in application/core/MY_Router.php and just copy the method _set_routing() in your class. Your changes would go somewhere after routes.php file is included and set to $this->routes property, you can add your custom routes.

include(APPPATH. config/routes .EXT);
...
$this->routes = ( ! isset($route) OR ! is_array($route)) ? array() : $route;
unset($route);

//Get your custom routes:
$your_routes = $this->_get_custom_routes();
foreach($your_routes as $custom_route)
{
    $this->routes[$custom_route[ your_regex_match ]] = $custom_route[ controller ]. / .$custom_route[ action ];
}

Of course you might not need this, but I hope it gives you some idea. This way you can add custom routes and since you will have to fetch them from database, consider caching them for better speed.

Answer 2

我感到,每星期都会问这个问题。

Open up your application/config/routes.php, then add something like this:

$route[ ^(:any) ] = "my_controller/get_article/$1";

Please note that it will route everything to a controller called action. If you have other controllers then you should add a route for them too (preferably placed before this one).

// EDIT: Using this you can goto http://your-site.com/secrets-of-internet-marketing and it will call the get_article function in the my_controller controller, and pass "secrets-of-internet-marketing" as the first argument. Which can then process with something like this:

public function get_article($article_name) { 
    // something like this: 
    $article = $this->article_model->get_model_by_name($article_name);
    $this->load->view( article , $article);
}

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