Re: How lo load a model from a model?

Put in a Behaviour and apply the behaviour to each model that needs to use it.

Jeremy Burns

Class Outfit

http://www.classoutfit.com

On 6 Mar 2012, at 11:37:03, heohni wrote:

No there is not.

But I just found out that I can use it this way:

$results = Objekt::translateStrings($results);

But - the function  translateStrings() is using inside a call to $this->ExtractUet() and now it happens that I get the message  

Fatal error: Call to undefined method DetailsController::ExtractUet()

What do I have to do? In order to use the complete  translateStrings functionality? 

Am Dienstag, 6. März 2012 12:17:28 UTC+1 schrieb Jeremy Burns:

Is there a model association between the two models? If so, you can daisy chain them:

$results = $this->Detail->Objekt->translateStrings($results);debug($results);

Jeremy Burns

Class Outfit

http://www.classoutfit.com

On 6 Mar 2012, at 11:14:42, heohni wrote:

Hi,

class DetailsController extends AppController{
var $name = 'Detail';

public function index($id){
if($id){
    $results = $this->Detail->find('first', $conditions['Gesamtobjekt']);

    App::uses('Objekt', 'Model');
    $results = $this->Objekt->translateStrings($results);debug($results);
}
}
}

What do I have to do, to get the  $this->Objekt->translateStrings working?
Right niow, I only get

Fatal error: Call to a member function translateStrings() 

Message.

What do I miss?

I tried to get an info from the book, but I only find how to write the App::uses but I didn't find any further note..

Please advice! 

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