CakePHP检索数据时忽略“hasMany”

我有两张桌子——盒子和苹果。苹果属于一个盒子,一个盒子里有许多苹果。中的代码如下所示:

class Apple extends AppModel {
    public $useDbConfig = 'somedb';
    public $useTable = 'apple';
    public $belongsTo = array(
        'Box' => array(
            'className' => 'Box',
            'foreignKey' => 'box_id'
        )
    );
} 
class Box extends AppModel {
    public $useDbConfig = 'somedb';
    public $useTable = 'box';
    public $hasMany = array(
        'Apple' => array(
            'className' => 'Apple',
            'foreignKey' => 'box_id',
            'dependent' => true
        )
    );
}

$this->Box->find("all")

(Array):
    [0] => (Array):
        [Box] => (Array):
            // some data
        [Apple] => (Array):
            [0] => (Array):
                // some data
            [1] => (Array):
                // some data
            ...
    [1] => (Array):
        [Box] => (Array):
            // some data
        [Apple] => (Array):
            [0] => (Array):
                // some data
            [1] => (Array):
                // some data
    ...

更新: 正如@Ben所回答的,我需要将recursive=-1(或0)放在这里。

Room $hasMany 房间 Box ,及 $belongsTo 选择 在里面 盒子

 class Room extends AppModel {
    public $useDbConfig = 'somedb';
    public $useTable = 'room';
    public $hasMany = array(
        'Box' => array(
            'className' => 'Box',
            'foreignKey' => 'room_id'
        )
    );
}
// Edited Box class
class Box extends AppModel {
    public $useDbConfig = 'somedb';
    public $useTable = 'box';
    public $hasMany = array(
        'Apple' => array(
            'className' => 'Apple',
            'foreignKey' => 'box_id',
            'dependent' => true
        )
    );
    public $belongsTo = array(
        'Room' => array(
            'className' => 'Room',
            'foreignKey' => 'room_id'
        )
    );
}

那么,我如何才能获得有关苹果、盒子和房间的数据呢?比方说,我有一个 房间 : where room.name like "room0%" . 我试了下一件事:

    $some = $this->Apple->find("all", array(
        'joins' => array(
            array('table' => 'room',
                'alias' => 'Room',
                'type' => 'INNER',
                'conditions' => array(
                    'Room.id = Box.room_id',
                )
            )
        ),
        'conditions' => array('Room.name LIKE' => 'room0%')
    ));

但我只收到一组苹果盒“对”。

目前我找到的唯一解决方案是next:

    $temp = "room0%";
    $db = $this->Apple->getDataSource();
    $some = $db->fetchAll(
        "SELECT * 
          FROM  apple 
          JOIN box on apple.box_id = box.id
          JOIN room on box.room_id = room.id
          WHERE room.name LIKE '$temp'"
    );

有没有一种方法可以用模型来做同样的事情?