thinkphp关联查询有关问题，join

thinkphp关联查询问题，join
$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 thinkphp 关联查询

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------解决方案--------------------
$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();