更新！
我对cakepp还是个新手，但对mysql和php有经验。模型看起来像：
Person->Father
父亲是自我指代的人。
我根据mysql查询写了以下内容，该查询返回了"1"个人的父亲mysql:
SELECT `Father`.name,`Father`.id from persons as `Father` left join persons as `Person` on `Person`.`father_id`=`Father`.`id` where `Person`.id=1
结块
$options = array(
'fields' => array(
//'Father.name',
'Father.id',
),
'joins' => array(
array(
'conditions' => array(
'Person.father_id = Father.id',
),
'table' => 'persons',
// 'alias' => 'Person', i commented because having conflict with scaffolded model
'type' => 'left',
),
),
'conditions' => array(
'Person.id' => '1',
),
'contain' => array(
'Person','Father',
),
);
$data = $this->Person->find('first', $options);
$fatherquery=$this->Person->find('first',array('conditions'=>array('Person.id'=>$data['Father']['id'])));
为了获得与mysql相同的结果，我添加了这一行（最后一行$father=…，但现在它似乎是子查询，看起来连接不起作用），因为father和Person不是同一个模型，如果我有
$data['Father']['name'] and $data['Person']['name'] they are not equal
顺便说一句，我已经有了解决方案，但也许我误解了一些概念。有没有办法让mysql查询更容易？
试试这个。（请注意缺少contain。使用contain的唯一原因是，如果您还没有在主模型的find（）或join中获得数据）：
//Person model (cake 2.x code)
$this->find('first', array(
'fields' => array(
'Father.id',
'Father.name'
),
'conditions' => array(
'Person.id' => 1
),
'joins' => array(
array(
'table' => 'persons',
'alias' => 'Father',
'type' => 'left',
'conditions' => array(
'Person.father_id = Father.id'
)
)
)
));