MySQL高阶1919-兴趣相同的朋友

题目

请写一段SQL查询获取到兴趣相同的朋友。用户 x 和用户 y 是兴趣相同的朋友，需满足下述条件：

用户 x 和 y 是朋友，并且
用户 x and y 在同一天内听过相同的歌曲，且数量大于等于三首.

结果表 无需排序 。注意：返回的结果需要和源数据表的呈现方式相同（例如，需满足 user1_id < user2_id）。

准备数据

Create table If Not Exists Listens (user_id int, song_id int, day date)
Create table If Not Exists Friendship (user1_id int, user2_id int)Truncate table Listensinsert into Listens (user_id, song_id, day) values ('1', '10', '2021-03-15')insert into Listens (user_id, song_id, day) values ('1', '11', '2021-03-15')insert into Listens (user_id, song_id, day) values ('1', '12', '2021-03-15')insert into Listens (user_id, song_id, day) values ('2', '10', '2021-03-15')insert into Listens (user_id, song_id, day) values ('2', '11', '2021-03-15')insert into Listens (user_id, song_id, day) values ('2', '12', '2021-03-15')insert into Listens (user_id, song_id, day) values ('3', '10', '2021-03-15')insert into Listens (user_id, song_id, day) values ('3', '11', '2021-03-15')insert into Listens (user_id, song_id, day) values ('3', '12', '2021-03-15')insert into Listens (user_id, song_id, day) values ('4', '10', '2021-03-15')insert into Listens (user_id, song_id, day) values ('4', '11', '2021-03-15')insert into Listens (user_id, song_id, day) values ('4', '13', '2021-03-15')insert into Listens (user_id, song_id, day) values ('5', '10', '2021-03-16')insert into Listens (user_id, song_id, day) values ('5', '11', '2021-03-16')insert into Listens (user_id, song_id, day) values ('5', '12', '2021-03-16')Truncate table Friendshipinsert into Friendship (user1_id, user2_id) values ('1', '2')insert into Friendship (user1_id, user2_id) values ('2', '4')insert into Friendship (user1_id, user2_id) values ('2', '5')

listens表

friendship表

分析数据

select distinct user1_id,user2_id
from Friendship fleft join Listens l1on user1_id=l1.user_idleft join Listens l2on user2_id=l2.user_id
where l1.song_id=l2.song_idand l1.day=l2.day
group by user1_id,user2_id,l1.day
having count(distinct l2.song_id)>=3;