Re: Get a model object instead of an array when calling $this->Model->find(id)

Hrm, using this it works:

$this->User->read(null, $this->params['named']['user_id']);
$this->User->getIni();

But I often wondered why CakePHP relies on this "proxy" or something
for its models? It seems I always have to communicate through
$this->Modelname when doing stuff with models, is it creating,
viewing, updating or deleting. Why is that? Why don't I work with
"real" dedicated objects? I guess there's a good reason (and I think
it's because of PHP, because in Ruby on Rails this is done quite more
elegant as far as I remember), but I couldn't find it yet.

On Wed, Oct 27, 2010 at 1:53 PM, psybear83 <psybear83@gmail.com> wrote:
> Hi everybody
>
> This seems pretty basic, but I just don't get it to work...
>
> I have a model User with a method:
>
>        function getIni() {
>          return "some INI stuff";
>        }
>
> Then I have an action downloadIni() in UsersController:
>
>  function downloadIni() {
>    // ...
>    $user = $this->User->find($this->params['named']['user_id']);
>    $user->getIni();
>    // ...
>  }
>
> However, because find() actually doesn't return a User model instance
> (but an array), this doesn't work. But sadly I just didn't find a way
> to get a model instance from $this->User...
>
> I found this:
> http://debuggable.com/posts/how-to-properly-create-a-model-instance-manually:480f4dd6-4424-4c89-9564-4647cbdd56cb
> But this seems to be outdated...
>
> Thanks for help!
>
> Check out the new CakePHP Questions site http://cakeqs.org and help others with their CakePHP related questions.
>
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Check out the new CakePHP Questions site http://cakeqs.org and help others with their CakePHP related questions.

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