Question

演员/电影查询有很多变化。当我尝试从非组列中进行选择时，我无法找到正确的方法：我有3个表电影（film_id，标题），演员（actor_id，first_name，last_name），film_actor（actor_id，film_id）。 所以，我需要找到2个演员，这些演员一起演出最多，并为前5部这样的电影显示输出为actor1_last_name，actor2_last_name，film_title。 我首先要做的是，算上演员的外表：

select r1.actor_id as actor_a, r2.actor_id as actor_b,
count(r1.film_id) as casted_together
from film_actor r1 inner join film_actor r2 on r1.film_id = r2.film_id
and r1.actor_id > r2.actor_id
group by r1.actor_id, r2.actor_id
order by casted_together desc

这会返回像

这样的表格

actor_a | actor_b| casted_together
Name      Name      7
...       ...       6

但是一旦我尝试添加film.title来选择它，我说如果它不在组部分（(select title from film where film_id = r1.film_id) as film_title）中我就不能使用它。 subquery uses ungrouped column "r1.film_id" from outer query

如何使用上述分组的结果并添加一些外部列？就我而言，它是film.title。

期望的输出：

actor_a_last_name | actor_b_last_name| film_title
Name                Name                Title
...                 ...                 ...

说，乔治克鲁尼和面包皮特的电影数量最多（9）结果应显示9行中的前5行

    actor_a_last_name | actor_b_last_name| film_title
     Clooney                Pitt            Film 1
     Clooney                Pitt            Film 2
     ...                    ...             ...

Answer 1

使用聚合，例如string_agg()或array_agg()：

select 
    r1.actor_id as actor_a, 
    r2.actor_id as actor_b,
    count(r1.film_id) as casted_together,
    string_agg(f.title, ', ') as films
from film_actor r1 
join film_actor r2 
    on r1.film_id = r2.film_id
    and r1.actor_id > r2.actor_id
join film f
    on r1.film_id = f.film_id
group by r1.actor_id, r2.actor_id
order by casted_together desc

将查询用作派生表并添加演员姓名和电影片名：

select 
    a1.last_name,
    a2.last_name,
    f.title
from (
    select 
        r1.actor_id as actor_a_id, 
        r2.actor_id as actor_b_id,
        count(r1.film_id) as casted_together,
        array_agg(r1.film_id) as film_ids
    from film_actor r1 
    join film_actor r2 
        on r1.film_id = r2.film_id
        and r1.actor_id > r2.actor_id
    join film f
        on r1.film_id = f.film_id
    group by r1.actor_id, r2.actor_id
    order by casted_together desc
    limit 1
    ) s
join actor a1 on a1.actor_id = s.actor_a_id
join actor a2 on a2.actor_id = s.actor_b_id
join film f on f.film_id = any(s.film_ids)

sql选择演员一起玩

1 个答案: