Have you got an example of that?
On 24 May 2013, at 01:09, lowpass <zijn.digi...@gmail.com> wrote:
> 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 <lut...@swbell.net> 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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