SQL 经典面试题：统计最近七天连续三天活跃的用户-Toy模板网

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1 需求

给定 mid，dt 的用户登录记录表，查找最近 7 天内连续 3 天活跃的用户 id

2 数据表

tmp_table.tmp_login_test

CREATE TABLE tmp_table.tmp_login_test (
    mid string,
    dt string
) ROW FORMAT DELIMITED FIELDS TERMINATED BY ',';

insert into tmp_table.tmp_login_test values
("zhangsan","2021-08-01"),
("zhangsan","2021-08-02"),
("zhangsan","2021-08-04"),
("zhangsan","2021-08-05"),
("zhangsan","2021-08-06"),
("zhangsan","2021-08-08"),
("zhangsan","2021-08-09"),
("zhangsan","2021-08-10"),
("lisi","2021-08-01"),
("lisi","2021-08-02"),
("lisi","2021-08-04"),
("lisi","2021-08-05"),
("lisi","2021-08-08"),
("lisi","2021-08-10"),
("wangwu","2021-08-01"),
("wangwu","2021-08-04"),
("wangwu","2021-08-07"),
("zhaoliu","2021-08-01"),
("zhaoliu","2021-08-02"),
("zhaoliu","2021-08-03"),
("zhaoliu","2021-08-04"),
("zhaoliu","2021-08-05"),
("zhaoliu","2021-08-06"),
("zhaoliu","2021-08-07"),
("zhaoliu","2021-08-08"),
("zhaoliu","2021-08-09"),
("zhaoliu","2021-08-10");

3 Sql 实现

实现思路：获取最近 7 天的用户登录记录数据（在 where 中限定），对数据进行 rank 排序，计算登录日期与 rank 值之间的差值（使用 date_sub 函数）得到一个差值日期，如果登录日期是连续的那么计算得到的差值日期是同一个，在此基础上基于用户，差值日期分组，统计 dt 的去重数量，即可得到每个用户每次连续登录的天数。在本例中，需要统计 7 天内连续 3 天登录的用户，所以只需要取出连续登录天数大于等于 3 的 uid 即完成需求。文章来源地址https://www.toymoban.com/news/detail-519750.html

对用户的登录行为按 mid 分组，组内按登录日期进行排序

select 
    mid, dt, 
    rank() over(partition by mid order by dt) rank_mid_dt 
from tmp_table.tmp_login_test 
where dt >= date_sub('2021-08-10', 6) and dt <= '2021-08-10';

+-----------+-------------+--------------+
|    mid    |     dt      | rank_mid_dt  |
+-----------+-------------+--------------+
| lisi      | 2021-08-04  | 1            |
| lisi      | 2021-08-05  | 2            |
| lisi      | 2021-08-08  | 3            |
| lisi      | 2021-08-10  | 4            |
| zhangsan  | 2021-08-04  | 1            |
| zhangsan  | 2021-08-05  | 2            |
| zhangsan  | 2021-08-06  | 3            |
| zhangsan  | 2021-08-08  | 4            |
| zhangsan  | 2021-08-09  | 5            |
| zhangsan  | 2021-08-10  | 6            |
| zhaoliu   | 2021-08-04  | 1            |
| zhaoliu   | 2021-08-05  | 2            |
| zhaoliu   | 2021-08-06  | 3            |
| zhaoliu   | 2021-08-07  | 4            |
| zhaoliu   | 2021-08-08  | 5            |
| zhaoliu   | 2021-08-09  | 6            |
| zhaoliu   | 2021-08-10  | 7            |
| wangwu    | 2021-08-04  | 1            |
| wangwu    | 2021-08-07  | 2            |
+-----------+-------------+--------------+

基于上表计算 dt 和 rank_mid_dt 日期差

select 
    mid, dt, rank_mid_dt, 
    date_sub(dt, rank_mid_dt) date_diff 
from (select mid, dt, rank() over(partition by mid order by dt) rank_mid_dt from tmp_table.tmp_login_test where dt >= date_sub('2021-08-10', 6) and dt <= '2021-08-10') t1;

+-----------+-------------+--------------+-------------+
|    mid    |     dt      | rank_mid_dt  |  date_diff  |
+-----------+-------------+--------------+-------------+
| lisi      | 2021-08-04  | 1            | 2021-08-03  |
| lisi      | 2021-08-05  | 2            | 2021-08-03  |
| lisi      | 2021-08-08  | 3            | 2021-08-05  |
| lisi      | 2021-08-10  | 4            | 2021-08-06  |
| zhangsan  | 2021-08-04  | 1            | 2021-08-03  |
| zhangsan  | 2021-08-05  | 2            | 2021-08-03  |
| zhangsan  | 2021-08-06  | 3            | 2021-08-03  |
| zhangsan  | 2021-08-08  | 4            | 2021-08-04  |
| zhangsan  | 2021-08-09  | 5            | 2021-08-04  |
| zhangsan  | 2021-08-10  | 6            | 2021-08-04  |
| zhaoliu   | 2021-08-04  | 1            | 2021-08-03  |
| zhaoliu   | 2021-08-05  | 2            | 2021-08-03  |
| zhaoliu   | 2021-08-06  | 3            | 2021-08-03  |
| zhaoliu   | 2021-08-07  | 4            | 2021-08-03  |
| zhaoliu   | 2021-08-08  | 5            | 2021-08-03  |
| zhaoliu   | 2021-08-09  | 6            | 2021-08-03  |
| zhaoliu   | 2021-08-10  | 7            | 2021-08-03  |
| wangwu    | 2021-08-04  | 1            | 2021-08-03  |
| wangwu    | 2021-08-07  | 2            | 2021-08-05  |
+-----------+-------------+--------------+-------------+

基于 mid，date_diff 分组，统计 dt 的去重数量，并取出数量大于 3 的

select 
    mid, date_diff, 
    count(distinct dt) cnt 
from (select mid, dt, rank_mid_dt, date_sub(dt, rank_mid_dt) date_diff from (select mid, dt, rank() over(partition by mid order by dt) rank_mid_dt from tmp_table.tmp_login_test where dt >= date_sub('2021-08-10', 6) and dt <= '2021-08-10') t1) t2 
group by mid, date_diff having count(distinct dt) >= 3;

+-----------+-------------+------+
|    mid    |  date_diff  | cnt  |
+-----------+-------------+------+
| zhaoliu   | 2021-08-03  | 7    |
| zhangsan  | 2021-08-03  | 3    |
| zhangsan  | 2021-08-04  | 3    |
+-----------+-------------+------+

取出 mid

select 
    distinct mid
from (select mid, date_diff, count(distinct dt) cnt from (select mid, dt, rank_mid_dt, date_sub(dt, rank_mid_dt) date_diff from (select mid, dt, rank() over(partition by mid order by dt) rank_mid_dt from tmp_table.tmp_login_test where dt >= date_sub('2021-08-10', 6) and dt <= '2021-08-10') t1) t2 
group by mid, date_diff having count(distinct dt) >= 3) t3;

+-----------+
|    mid    |
+-----------+
| zhangsan  |
| zhaoliu   |
+-----------+

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SQL 经典面试题：统计最近七天连续三天活跃的用户

1 需求

2 数据表

3 Sql 实现

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