Apply the array to the $paginate class var, not the method. Basically, you
can declare the $paginate array as a class var, then add to or override any
part of it from within an action.


On Thu, May 23, 2013 at 7:12 PM, Larry Lutz <[email protected]> wrote:

> I'm trying to achieve the fat model/skinny controller mantra in CakePHP
> 2.7. In that scenario, the model should control the data, and that means
> moving the finds out of the controllers and into the models. However, at
> least with the view and admin_view actions, one needs the data paginated.
>
> The problem starts there. It's easy enough to do a $this->find('all') in
> the model to get the data, along with any conditions, contains, etc.
> However that produces an array. Paginator->paginate(), as far as I can
> tell won't work in the model; it's a component that only works in a
> controller. Yet, if I take the array produced in the model into the
> controller and try to pass it on to Paginator->paginate(), it fails
> because paginate won't accept an array.
>
> In terms of trying to achieve the fat model/skinny controller goal, what
> am I missing here. How can I achieve a paginated result while still having
> the model (properly) do the work of managing and finding the data?
>
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