1.select * from film where film_id in ( select film_id from film.actor where actor_id = 1);
因?yàn)閕n后面的數(shù)據(jù)多了的話，會(huì)很影響效率
in() 加子查詢改進(jìn)
優(yōu)化修改：
2.select film.* from film inner join film_actor using(film_id) where actor_id = 1;
3.select * from film where exists (select * from film_actor where actor_id = 1 and film_actor.film_id = film.film_id);