我先从比尔·卡温(Zend_DB的候选者)那里读过一些与数据库表没有直接联系的模式。 (我认为,有些发展中国家有其模式,可直接延伸Zentd_tables,从而更难以增加 objects子的 objects子,这是有意义的。)

So what Bill Karwin was saying is that a Model has tables and isn t a table but I m still thinking the way I have it is correct as its designed in an object oriented manner.

例如(理由说明):

A Monster has 1:M Mouth. a Mouth has 1:M Tooth. 

因此,在我的数据库中,有5个表格:

Monster: id, name, type
MonsterMouth: id, monster_id, mouth_id
Mouth: id, size
MouthTeeth: id, mouth_id, tooth_id
Tooth: id, size, shape, sharpness

然后是3个班级:

class Model_Monster { 
    private $id, $name, $type, $mouths = array();
    public function __construct ($id) {
        // Set properties from DB for supplied ID
        // Go through DB and add the mouths based on monster ID
    }

    public function countTeeth () {
        // Loop through each $mouths as $mouth and call $mouth->getTotalTeeth();
    }
} 
class Model_MonsterMouth { 
    private $id, $size, $teeth = array(); 
    public function __construct($id) {
        // Set properties from DB for supplied ID
        // Go through DB and add the types of teeth for this mouth ID
    }

    public function getTotalTeeth () {
        // return sizeof($teeth);
    }

}
class Model_Tooth {
    private $id, $size, $shape, $sharpness;
    public function __construct($id) {
        // Populate details based on ID passed
    }
}

然后,我猜测计算牙齿和牙齿的方法。

$monsterId = 1;
$monster = new Monster($monsterId);
// Count total teeth
$totalTeeth = $monster->countTeeth();

因此,僧.可能有多种不同的口.,1口mouth可能有多种不同的牙齿。

After writing out this lengthy post I think I ve got it right and that Bill Karwin is talking about those who have 5 Models rather than 3...

页: 1

3个模型中的代码>2 用于储存许多其他类型的物体。

如果mon门在大约10种不同类型牙齿中的9-10克之间有10口mouth,这是否是一个业绩问题? 我指的是PHP。 如果一个物体有1-100k且在复合物品中添加1-100k,那么我就认为,我只应当拥有一个带有若干财产的物体,以表明有多少此类物品。

如果我听起来正确的话,也许会帮助一些与之有问题的其他人。

感谢:D Dom