Re: Using variable in hasmany condition

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 <zijn.digital@gmail.com> wrote:

On Wed, Nov 24, 2010 at 8:46 PM, Pedro <pedrohmello@gmail.com> 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.

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