数据库每日一题 --- 第 11 天：合作过至少三次的演员和导演

+-------------+---------+| Column Name | Type    |+-------------+---------+| actor_id    | int     || director_id | int     || timestamp   | int     |+-------------+---------+timestamp 是这张表的主键.

ActorDirector 表：+-------------+-------------+-------------+| actor_id    | director_id | timestamp   |+-------------+-------------+-------------+| 1           | 1           | 0           || 1           | 1           | 1           || 1           | 1           | 2           || 1           | 2           | 3           || 1           | 2           | 4           || 2           | 1           | 5           || 2           | 1           | 6           |+-------------+-------------+-------------+
Result 表：+-------------+-------------+| actor_id    | director_id |+-------------+-------------+| 1           | 1           |+-------------+-------------+唯一的 id 对是 (1, 1)，他们恰好合作了 3 次。

1.聚合函数2.建议用时10~25min

select actor_id, director_id,count(*) as 'number'from   ActorDirectorgroup by actor_id, director_id

{"headers": ["actor_id", "director_id", "number"], "values": [ [1, 1, 3],             [2, 1, 2],             [1, 2, 2]]}

select actor_id, director_idfrom   ActorDirectorgroup by actor_id, director_idhaving count(*)>=3