Re: Selecting users who are not friends

Thank you Arthur - I figured I'd have to do it that way - was just
looking for a more elegant solution.

Thanks!
~R~

On Dec 29, 1:54 pm, "Arthur Pemberton" <pem...@gmail.com> wrote:
> On Mon, Dec 29, 2008 at 2:59 PM, WebFeathers <webfeath...@gmail.com> wrote:
>
> > Hey-
> > I'm trying to generate a list of users who are not friends of the
> > currently logged in user.
> > I have two tables: profiles, and users_users
> > profiles includes the fields: id, first_name, last_name, etc...
> > users_users includes: id, user_id, friend_id
>
> > I can get the list of users who ARE friends:
> > $friends = $this->User->findAll('User.id=' . $target);
>
> > ...but can't figure out how to go the other way???
>
> > My user model includes:
> > class User extends AppModel
> > {
> >        var $name = 'User';
> >        var $hasAndBelongsToMany = array(
> >                'Friends' => array(
> >                        'className' => 'Profile',
> >                        'joinTable' =>'users_users',
> >                        'foreignKey' =>'user_id',
> >                        'associationForeignKey' => 'friend_id',
> >                        'unique'=> true
> >                        ),
> >                'FriendList' => array(
> >                        'className' => 'users_users',
> >                        'joinTable' =>'users_users',
> >                        'foreignKey' =>'user_id',
> >                        'associationForeignKey' => 'friend_id',
> >                        'unique'=> true
> >                        )
> >        );
> > }
>
> I believe if you did this in raw SQL, you would need to do a sub
> query. I don't think CakePHPs ORM generates sub queries (subject to
> correction).
>
> The simplest solution would be to generate a list of the ids of the
> friends, and get all users whose id is not in that list.
>
> [1] show how to do an IN condition
>
> [1]http://book.cakephp.org/view/74/Complex-Find-Conditions
>
> --
> Fedora 9 : sulphur is good for the skin
> (www.pembo13.com)
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