in your Property model $hasMany add:
var $hasMany = array('Picture' =>
array('className' => 'Picture',
'foreignKey' => 'property_id'),
'fields' => array('id', 'name'),
);
replacing 'id', and 'name' with whatever you would like returned.
As a note - its generally a bad idea to store photos in a database.
Just store their filename and location.
Hope that helps.
On Dec 13, 5:04 pm, Miguel XT <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Let's say I have two models Property and Picture. One Property can
> have many Pictures. So I make two models like this:
>
> class Property extends AppModel {
> var $name = "Property";
>
> var $hasMany = array('Picture' =>
> array('className' => 'Picture',
> 'foreignKey' => 'property_id')
> );
>
> }
>
> class Picture extends AppModel {
> var $name = "Picture";
>
> var $belongsTo = array('Property' =>
> array('className' => 'Property',
> 'conditions' => '',
> 'order' => '',
> 'foreignKey' => 'property_id'
> )
> );
>
> }
>
> Now I want to use $this->Property->findAll with some conditions and
> return ALL the Property model, but NOT ALL the picture Model. That is
> because the Picture model have large image data that I don't need. I
> just need the picture ID returned.
>
> What is the best way to accomplish this?
>
> Regards
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