thinkphp关联查询问题，join

thinkphp 关联查询 $result = $room->join('r_hospital on r_department.hospital_id=r_hospital.id')->where(array('hospital_id'=>array('exp','is not null')))->select();
大神们看看，where(array('hospital_id'=>array('exp','is not null')))这句话是什么意思？结果显示出来所有的医院，但我只想查某一个，把医院id等于$data，怎么做

回复讨论(解决方案)

没人会吗？难到就那么难吗

从字面理解是hostpital_id中非NULL空的都选择

$condition['hospital_id'] = $data;
// 把查询条件传入查询方法
$result = $room->join('r_hospital on r_department.hospital_id=r_hospital.id')->where($condition)->select();

$condition['hospital_id'] = $data;
// 把查询条件传入查询方法
$result = $room->join('r_hospital on r_department.hospital_id=r_hospital.id')->where($condition)->select();
恩，确实是这种方法。三级关联的怎么写，再添加一个医生doctor的id

$condition['hospital_id'] = $data;
// 把查询条件传入查询方法
$result = $room->join('left join r_hospital on r_department.hospital_id=r_hospital.id left join doctor on doctor.id = xx.id')->where($condition)->select();

$condition['hospital_id'] = $data;
// 把查询条件传入查询方法
$result = $room->join('left join r_hospital on r_department.hospital_id=r_hospital.id left join doctor on doctor.id = xx.id')->where($condition)->select();
谢了

$condition['hospital_id'] = $data;
// 把查询条件传入查询方法
$result = $room->join('left join r_hospital on r_department.hospital_id=r_hospital.id left join doctor on doctor.id = xx.id')->where($condition)->select();
这个是三级关联的吗？貌似不行呀。帮我写个三级关联的吧，医生属于科室，科室属于医院这种关系。我弄了很久了，就是不会

请帖出3张表结构