Hi,
You can also choose to write relation's conditions in app model's
constructor, something like
$this->hasMany['Log']['conditions'] = array('Log.model' => $this->name);
Amit Badkas
PHP Applications for E-Biz: http://www.sanisoft.com
On Thu, Nov 25, 2010 at 8:57 AM, cricket <[email protected]> wrote:
> On Wed, Nov 24, 2010 at 8:46 PM, Pedro <[email protected]> wrote:
> > class AppModel extends Model
> > {
> >
> > var $hasMany = array
> > (
> > 'Log' => array
> > (
> > 'className' => 'IsaLog',
> > 'foreignKey' => 'parent_id',
> > 'conditions' => array('Log.model' => $this->name),
> > 'order' => 'Log.inserted_at'
> > )
> > );
> > }
> >
> > generates
> > Parse error: syntax error, unexpected T_VARIABLE in on line 43
> >
> > It possible to do something like that??
>
> Nope. You can't use variables when declaring class members. It looks
> like you'd be better off creating a behavior for this that added Log
> to the model's $hasMany array.
>
> 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
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