目录
- [1 题目](#1 题目)
- [2 建表语句](#2 建表语句)
- [3 题解](#3 题解)
题目来源：腾讯。
1 题目
有两个表，朋友关系表user_friend，用户步数表user_steps。朋友关系表包含两个字段，用户id，用户好友的id；用户步数表包含两个字段，用户id，用户的步数.用户在好友中的排名
-- user_friend 数据
+----------+------------+
| user_id | friend_id |
+----------+------------+
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 3 |
| 2 | 4 |
| 2 | 5 |
| 3 | 1 |
| 3 | 4 |
| 3 | 5 |
| 4 | 2 |
| 4 | 3 |
| 4 | 5 |
| 5 | 2 |
| 5 | 3 |
| 5 | 4 |
+----------+------------+
--user_friend数据
+---------------------+-------------------+
| user_steps.user_id | user_steps.steps |
+---------------------+-------------------+
| 1 | 100 |
| 2 | 95 |
| 3 | 90 |
| 4 | 80 |
| 5 | 10 |
+---------------------+-------------------+
2 建表语句
sql
CREATE TABLE user_friend
(
user_id INT,
friend_id INT
) ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t';
-- 插入数据
INSERT INTO user_friend
VALUES (1, 2),
(1, 3),
(2, 1),
(2, 3),
(2, 4),
(2, 5),
(3, 1),
(3, 4),
(3, 5),
(4, 2),
(4, 3),
(4, 5),
(5, 2),
(5, 3),
(5, 4);
CREATE TABLE user_steps
(
user_id INT,
steps INT
) ROW FORMAT DELIMITED FIELDS TERMINATED BY '\t';
INSERT INTO user_steps
VALUES (1, 100),
(2, 95),
(3, 90),
(4, 80),
(5, 10);3 题解
（1）列出好友步数，并将自己步数添加到结果中
sql
--好友步数
select t1.user_id, t1.friend_id, t2.steps
from user_friend t1
join user_steps t2
on t1.friend_id = t2.user_id
union all
-- 自己步数
select user_id, user_id as friend_id, steps
from user_steps执行结果
+--------------+----------------+------------+
| _u1.user_id | _u1.friend_id | _u1.steps |
+--------------+----------------+------------+
| 1 | 2 | 95 |
| 1 | 3 | 90 |
| 2 | 1 | 100 |
| 2 | 3 | 90 |
| 2 | 4 | 80 |
| 2 | 5 | 10 |
| 3 | 1 | 100 |
| 3 | 4 | 80 |
| 3 | 5 | 10 |
| 4 | 2 | 95 |
| 4 | 3 | 90 |
| 4 | 5 | 10 |
| 5 | 2 | 95 |
| 5 | 3 | 90 |
| 5 | 4 | 80 |
| 1 | 1 | 100 |
| 2 | 2 | 95 |
| 3 | 3 | 90 |
| 4 | 4 | 80 |
| 5 | 5 | 10 |
+--------------+----------------+------------+
（2）按照用户分组，给每个用户的"好友"进行排名
sql
select tt1.user_id,
tt1.friend_id,
tt1.steps,
row_number() over (partition by tt1.user_id order by tt1.steps desc) as row_num
from (
--好友步数
select t1.user_id,
t1.friend_id,
t2.steps
from user_friend t1
join user_steps t2
on t1.friend_id = t2.user_id
union all
-- 自己步数
select user_id,
user_id as friend_id,
steps
from user_steps) tt1执行结果
+--------------+----------------+------------+----------+
| tt1.user_id | tt1.friend_id | tt1.steps | row_num |
+--------------+----------------+------------+----------+
| 1 | 1 | 100 | 1 |
| 1 | 2 | 95 | 2 |
| 1 | 3 | 90 | 3 |
| 2 | 1 | 100 | 1 |
| 2 | 2 | 95 | 2 |
| 2 | 3 | 90 | 3 |
| 2 | 4 | 80 | 4 |
| 2 | 5 | 10 | 5 |
| 3 | 1 | 100 | 1 |
| 3 | 3 | 90 | 2 |
| 3 | 4 | 80 | 3 |
| 3 | 5 | 10 | 4 |
| 4 | 2 | 95 | 1 |
| 4 | 3 | 90 | 2 |
| 4 | 4 | 80 | 3 |
| 4 | 5 | 10 | 4 |
| 5 | 2 | 95 | 1 |
| 5 | 3 | 90 | 2 |
| 5 | 4 | 80 | 3 |
| 5 | 5 | 10 | 4 |
+--------------+----------------+------------+----------+
（3）求取最终结果
sql
select user_id,
row_num
from (select tt1.user_id,
tt1.friend_id,
tt1.steps,
row_number() over (partition by tt1.user_id order by tt1.steps desc) as row_num
from (
--好友步数
select t1.user_id,
t1.friend_id,
t2.steps
from user_friend t1
join user_steps t2
on t1.friend_id = t2.user_id
union all
-- 自己步数
select user_id,
user_id as friend_id,
steps
from user_steps) tt1) tt2
where user_id = friend_id执行结果
+----------+----------+
| user_id | row_num |
+----------+----------+
| 1 | 1 |
| 2 | 2 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
+----------+----------+