Question

我有2个模型Scale和Step，其中model Scale有很多Step。 db表是：

scale : id | name | s1 | s2 | s3 | s4 | s5 | s6 | s7
step : id | val

我想在cakephp结构中编写以下查询...

select s.id, s1.val as v1, s2.val as v2, s3.val as v3, s4.val as v4, s5.val as v5, s6.val as v6, s7.val as v7
from scale s
join step s1 on s.s1 = s1.id
join step s2 on s.s2 = s2.id
join step s3 on s.s3 = s3.id
join step s4 on s.s4 = s4.id
join step s5 on s.s5 = s5.id
join step s6 on s.s6 = s6.id
join step s7 on s.s7 = s7.id

你能帮我解释一下语法吗？

Answer 1

我建议您改变数据库结构。而不是将Step绑定到S1，s2，s3 ......等Scale，这有一些缺点，其中之一就是它限制你每个Scale只有7个步骤，而且实际上并不是一个好的关系数据库模型，我建议您创建第三个表，称为ScaleSteps，并有三列：

id | ScaleId | StepId

假设您有一个ID为7的Scale条目，并且该Scale具有与之关联的Step条目1,2和3，结果表将如下所示：

id | ScaleId | StepId
1  | 7       | 1
1  | 7       | 2
1  | 7       | 3

然后使用：hasOne，belongsTo ...等设置CakePHP模型关系，例如：

class Scale extends Model {
var $name = "Scale";
var $hasMany = "ScaleStep"; 
}

class ScaleStep extends Model {
var $name = "ScaleStep";
var $belongsTo = "Step";    
}

Answer 2

$this->find('all', array(

        'fields' => array('s1.val', 's2.val'),
        'joins' => array(array(
            'table' => 'step',
            'alias' => 's1',
            'type' => 'inner',
            'conditions' =>array('Scale.s1 = s1.id')
            ),array(
            'table' => 'step',
            'alias' => 's2',
            'type' => 'inner',
            'conditions' =>array('Scale.s2 = s2.id')
            ),
        ),
        'conditions' => array('Scale.id = '.$id.'')
        ));

cakephp将查询转换为cakephp模型查询

2 个答案: