Why not just use a single controller that $uses each of the models in question?
Router::connect(
'/reports/:type',
array(
'controller' => 'reports',
'action' => 'view'
),
array(
'type' => '[a-z]+',
'pass' => array('type')
)
);
public function view($type = null)
{
if (!$type)
{
// ...
}
$model = Inflector::modelize($type);
// ...
}
On Tue, Oct 12, 2010 at 7:27 PM, Raisen <[email protected]> wrote:
> Assume that I have a reports page where there would be multiple sub-
> reports. I want the urls to look like:
>
> /reports/sales/
> /reports/customers/
>
> One way I managed to do that was to create actions inside the reports
> controller for each subreport, but I am trying to do something
> different. I want each subreport to have its own controller,view.
> Basically I would have a controller named:
>
> reports_sales_controller
>
> and views
>
> /reports/sales/index.ctp
>
> Is that possible?
>
> I was trying to do something like:
>
> Router::connect('/reports/(.*)/*', array('controller'=>'$1',
> 'action'=>'index'));
>
> which probably it's way off what I want.
>
> Check out the new CakePHP Questions site http://cakeqs.org and help others
> with their CakePHP related questions.
>
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