thinkphp關聯查詢有關問題，join

thinkphp關聯查詢問題，join
$result = $room->join('r_hospital on r_department.hospital_id=r_hospital.id')->where(array('hospital_id'=>array('exp','is not null')))->select();
大神們看看，where(array('hospital_id'=>array('exp','is not null')))這句話是什麼意思？結果顯示出來所有的醫院，但我只想查某一個，把醫院id等於$data，怎麼做 thinkphp 關聯查詢

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------解決方案--------------------
$condition['hospital_id'] = $data;
// 把查詢條件傳入查詢方法
$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();

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