如何获取使我的陈述中的陈述完整的值？

Question

我尝试展示每对演员，这两个演员没有在任何共同的电影类型中演出，而同时一个人和另一个人一起演出的类型至少有7个

我这样做：

select a1.actor_id as i8opoios1,a2.actor_id as i8opoios2,((count(distinct(g1.genre_name))+count(distinct(g2.genre_name)))>=7) as result from actor as a1
inner join actor as a2 on a1.actor_id!=a2.actor_id

inner join role as r1 on a1.actor_id=r1.actor_id
inner join movie as m1 on m1.movie_id=r1.movie_id
inner join movie_has_genre as mg1 on mg1.movie_id=m1.movie_id
inner join genre as g1 on mg1.genre_id=g1.genre_id

inner join role as r2 on a2.actor_id=r2.actor_id
inner join movie as m2 on m2.movie_id=r2.movie_id
inner join movie_has_genre as mg2 on mg2.movie_id=m2.movie_id
inner join genre as g2 on mg2.genre_id=g2.genre_id

where a1.actor_id<a2.actor_id and mg1.genre_id!=mg2.genre_id 
group by a1.actor_id,a2.actor_id;

该查询返回所有未曾在任何普通电影流派上演出的演员对，如果将它们组合成7种或以上流派，则返回1（TRUE），如果没有，则为0（FAlSE）。我的问题是是否有人有一个想法，我怎么能只返回真实的陈述。

Tables and their columns:

actor(actor_id,name)
role(actor_id,movie_id)
movie(movie_id,title)
movie_has_genre(movie_id,genre_id)
genre(genre_id,gender_name)

Answer 1

将条件添加到where子句以限制行。

SELECT 
    a1.actor_id as i8opoios1,
    a2.actor_id as i8opoios2,
    IF((count(distinct(g1.genre_name))+count(distinct(g2.genre_name)))>=7,1,0) as result 
FROM actor as a1
INNER JOIN actor as a2 
    on a1.actor_id != a2.actor_id
INNER JOIN role as r1 
    on a1.actor_id = r1.actor_id
INNER JOIN movie as m1 
    on m1.movie_id = r1.movie_id
INNER JOIN movie_has_genre as mg1 
    on mg1.movie_id = m1.movie_id
INNER JOIN genre as g1 
    on mg1.genre_id = g1.genre_id
INNER JOIN role as r2 
    on a2.actor_id = r2.actor_id
INNER JOIN movie as m2 
    on m2.movie_id = r2.movie_id
INNER JOIN movie_has_genre as mg2 
    on mg2.movie_id = m2.movie_id
INNER JOIN genre as g2 
    on mg2.genre_id = g2.genre_id
WHERE a1.actor_id < a2.actor_id 
    AND mg1.genre_id != mg2.genre_id 
HAVING IF((count(distinct(g1.genre_name))+count(distinct(g2.genre_name)))>=7,1,0) = 1
GROUP BY a1.actor_id,a2.actor_id;