Kohana: Requested URL was Not Found - php

Question

So I'm trying to install and get running Kohana. I am very new to it and frameworks in general (though I have used CakePHP a little bit).
Anyways...in my bootstrap file I have this:
// GET PARAMS -- This basically splits domain.com/kohana/controller/action/param1/etc
// into: controller | action | param1 | etc
$requestURI = explode('/', $_SERVER['REQUEST_URI']);
$scriptName = explode('/',$_SERVER['SCRIPT_NAME']);
for($i= 0;$i < sizeof($scriptName);$i++){
if ($requestURI[$i] == $scriptName[$i]){
unset($requestURI[$i]);
}
}
$param = array_values($requestURI);
$controller = # $param[0];
$action = # $param[1];
$param1 = # $param[2];
$param2 = # $param[3];
$param3 = # $param[4];
$param4 = # $param[5];
$param5 = # $param[6];
$param6 = # $param[7];
Now, I want to connect to my database and basically see if the first param aka $param[0] aka $controller is equal to one of my pages "categories" (corresponds to my categories_pages table). If it IS a category then I want to use the default "categories" controller otherwise, the controller should be whatever $param[0] is. This means if I go to domain.com/kohana/movies or domain.com/kohana/games it will display the categories controller otherwise domain.com/kohana/users will display the users controller.
$db = Database::instance();
$getiscategory = DB::select('*')->from('categories_pages')->where('directory', '=', $controller)->execute();
$is_category = $getiscategory->count();
if($is_category){
$controller = "categories";
}
$controller = (empty($controller)) ? 'index' : $controller;
$action = (empty($action)) ? 'index' : $action;
What I wanted to accomplish above works. If I echo $is_category I see that the value 1 is returned when there is a category match and if I echo $controller, I see that "categories" is set to be the controller.
Here is my code for the actual routing method...
Route::set(
'custom',
'(<controller>(/<action>(/<param1>)(/<param2>(/<param3>(/<param4>(/<param5>(/<param6>)))))))'
)->defaults(array(
'controller' => $controller,
'action' => $action,
'param1' => $param1,
'param2' => $param2,
'param3' => $param3,
'param4' => $param4,
'param5' => $param5,
'param6' => $param6,
));
Unfortunately, I'm not sure where it's routing to. As I mentioned $controller is returned previously as categories which is correct but yet I receive the error message "HTTP_Exception_404 [404]: The Requested URL $param[0] (movies or games, etc) was not found on this server."
Keep in mind I do not have a controller class for $param[0] if it matches a category because I want to use the "categories" controller class. If I go to domain.com/kohana/categories it works fine.
Anyone have any ideas/know a work-around?
Addition #1
I figured out that even though the categories controller is called and the correct action is called, it still is requiring the "shows" controller to display. I added a shows controller with the very basic info (template, content, etc) and it showed correctly. Is there a work around in the routing class to make the designated controller show? Like I said, I tell it what controller to go to and it acknowledges it but it doesn't actually go to it.

Answer 1

You're overcomplicating things I think. Also your code would force a limit of 6 parameters on every request.
I'll probably be easier to use two different routes, and if it doesn't match the first, then it can fall back to the second.
Route::set(
'categories',
'(<category>(/<action>(/<param1>)(/<param2>(/<param3>(/<param4>(/<param5>(/<param6>)))))))',
array('category' => '(movies|games)')
)->defaults(array(
'controller' => 'category',
'action' => 'index',
));
Route::set(
'users',
'<username>(/<action>(/<param1>)(/<param2>(/<param3>(/<param4>(/<param5>(/<param6>))))))'
)->defaults(array(
'controller' => 'users',
'action' => 'index',
));
If you only have one or two categories you could build them into the regex string, with some caching of course.
Otherwise, look up lambda routes, they're the means by which you can have dynamic routes like this.