php - 仅查找连接条件匹配的 parent

Question

我有两个相关的模型。现在我需要从 Field-Model 检索数据，其中 UserField 上的连接条件匹配:

我有 UserField 的 ID，想得到它的相关字段(-data): 相应的 sql 如下所示:

SELECT Field.* 
FROM fields AS Field 
  JOIN user_fields AS UserField 
    ON UserField.field_id = Field.id 
WHERE UserField.id = 3;

我想用蛋糕模型逻辑来写这个。

模型:

class Field extends AppModel {
     public $hasMany = array(
          'UserField' => array(
               'className' => 'UserField',
               'foreignKey' => 'field_id'
          )
     );
}

class UserField extends AppModel {
     public $belongsTo = array(
          'Field' => array(
               'className' => 'Field'
          )
     );
}

查找:

$this->Field->hasMany['UserField']['conditions'] = array('UserField.id' => 3);
$this->Field->find('all');

查询的

Result都是Field，而UserField只在条件匹配的地方设置。我需要将结果限制为仅匹配的情况:

array(
    (int) 0 => array(
        'Field' => array(
            'id' => (int) 1
        ),
        'UserField' => array()
    ),
    (int) 1 => array(
        'Field' => array(
            'id' => (int) 2
        ),
        'UserField' => array()
    ),
    (int) 2 => array(
        'Field' => array(
            'id' => (int) 3
        ),
        'UserField' => array(
            'id' => (int) 3
        )
    )
)

我的期望

array(
    (int) 2 => array(
        'Field' => array(
            'id' => (int) 3
        ),
        'UserField' => array(
            'id' => (int) 3
        )
    )
)

UserField 是可选的，我只想检索 Field 数据。

Answer 1

我相信你想做这样的事情:

$rows = $this->Field->find('all', array(
    'conditions' => array(
        'UserField.id' => 3
    )
);