you can use : $uses variable for include another model in your current
model.
$uses = array('modelnameA','modelnameB');
2009/5/11 Faza <f...@ansi.pl>
>
> Hi all,
>
> I've run into a weird problem.
>
> I have two models, not directly related, but I need data from model A to
> modify model B find results using ModelB::afterFind();
>
> In model B, i use this:
>
> $myModelA = ClassRegistry::init('ModelA');
>
> this returns no error.
>
> but when I do
> $values = $myModelA->findById($valid_id_from_somewhere)
>
> CakePHP fails to load page, but shows no errors.
>
> Anybody have a clue?
>
> thanks in advance...
> --
> Jacek
>
>
> >
>
--
Regards,
Mohammad Arif Hossen
Junior Web Developer
United Group International(UGIBD)
Mobile: +88 01714355911
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