Question

我有这张桌子：

actors(id: int, first_name: string, last_name: string, gender: string)
directors(id: int, first_name: string, last_name: string)
directors genres(director id: int, genre: string, prob: 
float)
movies(id: int, name: string, years: int, rank: 
float,)
movies directors(director id: int, movie id: int)
movies genres(movie id: int, genre: string)
roles(actor id: int, movie id: int, role: string)

我想找到每种类型的年份，其中制作了该类型的最大电影。

我正在做以下但是我被困住了，请帮忙！

select m.YEAR, count(m.year) as c, genre 
from movies_genres, 
     movies m 
where m.id = movies_genres.movie_id 
group by genre, m.year;

Answer 1

你每年都会得到每种类型的电影数量，这很棒。现在，您只需要将查询作为派生表来选择最大值。

select genre, year, max(c) mc
from
    (select m.YEAR, count(m.year) as c, genre 
    from movies_genres mg
         inner join movies m
             on m.id = mg.movie_id 
    group by genre, m.year)
group by genre, year

Answer 2

如果您使用的是SQL Server或Oracle或DB2，则应该可以使用。

SELECT genre, year, number
FROM 
(
  SELECT genre, year, number,
         row_number() over (PARTITION BY genre ORDER BY number DESC) as rank
  FROM
  (
    SELECT mg.genre, m.year, count(*) as number
    FROM movies_genres mg
    JOIN movies m on m.id = mg.movie_id
    GROUP BY mg.genre, m.year
  ) A
) B
WHERE rank = 1

它如何运作：从内到外，首先你得到所有类型和年份的计数。然后按计数对每个类型的年份进行排名，最后选择最大的项目。

如何使用group by编写sql查询

2 个答案: