mysql - CakePHP X hasMany Y, 找到有 Y.name = "value1" AND 同时有 Y.name = "value2" 的 X

Question

所以我的模型关联如下：

人有很多奖

人员列：id、name 奖励列：id、person_id、name

所以人们可以有很多奖项，但我只想检索那些同时获得两个不同奖项的特定人。有没有办法用 find 方法实现这一点？

Answer 1

$persons = $this->Person->find('all', array(
                           'recursive' => -1,
                           'conditions' => array(
                              'AND' => array('Award.name' => 'Award 1', 'Award.name' => 'Award 3')
                            ),
                           'joins' => array(
                                array(
                                   'table' =>'awards',
                                   'alias' => 'Award',
                                   'type' => 'LEFT',
                                   'conditions' => 'Award.person_id = person.id'
                                 )
                            )
                      )
                 );

Answer 2

查看页面末尾的子查询。您基本上需要一个子查询，因为您需要查询 Award 表两次以匹配 value1 和 value2。

---

编辑： 示例代码

$conditionsSubQuery['`Award`.`name`'] = 'value2';
$dbo = $this->Person->getDataSource();
$subQuery = $dbo->buildStatement(
    array(
        'fields' => array( '`Person2`.`id`'),
        'table' => $dbo->fullTableName($this->Person),
        'alias' => 'Person2',
        'limit' => null,
        'offset' => null,
        'joins' => array(),
        'conditions' => $conditionsSubQuery,
        'order' => null,
        'group' => null
    ),
    $this->Person
);
$subQuery = ' `Person`.`id` IN (' . $subQuery . ') ';
$subQueryExpression = $dbo->expression($subQuery);
$conditions[] = array('Award.name' => 'value1');
$conditions[] = $subQueryExpression;
$this->Person->find('all', compact('conditions'));

我没有测试它，它可能不起作用，但你明白了。

---

编辑2： 回答您的评论

所以基本上你想要的 SQL 结果是

select 
    users.id 
from 
    users 
left join 
    awards 
where 
    awards.name in ('award1', 'award2', 'award3', 'award4') 
group by 
    awards.name 
having 
    count(awards.name) = 4

为此，您可以这样做：

$awards = array('award1', 'award2', 'award3', 'award4');

$this->Person->find(
    'all',
    array(
        'joins' => array(
            array(
               'table' =>'awards',
               'alias' => 'Award',
               'type' => 'LEFT',
               'conditions' => 'Award.person_id = person.id'
             )
        ),
        'conditions' => array(
            'Award.name' => $awards,
        ),
        'group' => array('Award.name HAVING '.count($awards))
    )
);