目录

题目

准备数据

分析数据

实现

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题目

写出一个查询语句，找到具有最多共同关注者的所有两两结对组。换句话说，如果有两个用户的共同关注者是最大的，我们应该返回所有具有此最大值的两两结对组

结果返回表，每一行应该包含user1_id和 user2_id，其中user1_id < user2_id.

返回结果 不要求顺序 。

准备数据

Create table If Not Exists Relations (user_id int, follower_id int)
    Truncate table Relations
    insert into Relations (user_id, follower_id) values ('1', '3')
    insert into Relations (user_id, follower_id) values ('2', '3')
    insert into Relations (user_id, follower_id) values ('7', '3')
    insert into Relations (user_id, follower_id) values ('1', '4')
    insert into Relations (user_id, follower_id) values ('2', '4')
    insert into Relations (user_id, follower_id) values ('7', '4')
    insert into Relations (user_id, follower_id) values ('1', '5')
    insert into Relations (user_id, follower_id) values ('2', '6')
    insert into Relations (user_id, follower_id) values ('7', '5')

分析数据

第一步:根据共同好友进行关联条件,并且确保a.user_id < b.user_id

select a.user_id as user1_id, b.user_id as user2_id, a.follower_id from Relations a
        join Relations b on a.follower_id = b.follower_id and a.user_id<b.user_id;

第二步:对共同好友统计个数

select user1_id, user2_id, count(*) as cnt from
    (
        select a.user_id as user1_id, b.user_id as user2_id, a.follower_id from Relations a
                                                                                    join Relations b on a.follower_id = b.follower_id and a.user_id<b.user_id
    ) as temp
group by user1_id, user2_id;

第三步:筛选出具有这个最大值的用户对

with temp as
(
    select user1_id, user2_id, count(*) as cnt from
    (
        select a.user_id as user1_id, b.user_id as user2_id, a.follower_id from Relations a
        join Relations b on a.follower_id = b.follower_id and a.user_id<b.user_id
    ) as temp
    group by user1_id, user2_id
)

select user1_id, user2_id from temp
where cnt = (select max(cnt) from temp);

实现

with temp as
(
    select user1_id, user2_id, count(*) as cnt from
    (
        select a.user_id as user1_id, b.user_id as user2_id, a.follower_id from Relations a
        join Relations b on a.follower_id = b.follower_id and a.user_id<b.user_id
    ) as temp
    group by user1_id, user2_id
)

select user1_id, user2_id from temp
where cnt = (select max(cnt) from temp);