0 问题描述

用户行为日志表tb_user_log

id	uid	artical_id	in_time	out_time	sign_in
1	101	0	2021-07-07 10:00:00	2021-07-07 10:00:09	1
2	101	0	2021-07-08 10:00:00	2021-07-08 10:00:09	1
3	101	0	2021-07-09 10:00:00	2021-07-09 10:00:42	1
4	101	0	2021-07-10 10:00:00	2021-07-10 10:00:09	1
5	101	0	2021-07-11 23:59:55	2021-07-11 23:59:59	1
6	101	0	2021-07-12 10:00:28	2021-07-12 10:00:50	1
7	101	0	2021-07-13 10:00:28	2021-07-13 10:00:50	1
8	102	0	2021-10-01 10:00:28	2021-10-01 10:00:50	1
9	102	0	2021-10-02 10:00:01	2021-10-02 10:01:50	1
10	102	0	2021-10-03 10:00:55	2021-10-03 11:00:59	1
11	102	0	2021-10-04 10:00:45	2021-10-04 11:00:55	0
12	102	0	2021-10-05 10:00:53	2021-10-05 11:00:59	1
13	102	0	2021-10-06 10:00:45	2021-10-06 11:00:55	1

（uid-用户ID, artical_id-文章ID, in_time-进入时间, out_time-离开时间, sign_in-是否签到）

场景逻辑说明：

artical_id-文章ID代表用户浏览的文章的ID，特殊情况artical_id-文章ID为0表示用户在非文章内容页（比如App内的列表页、活动页等）。注意：只有artical_id为0时sign_in值才有效。
从2021年7月7日0点开始，用户每天签到可以领1金币，并可以开始累积签到天数，连续签到的第3、7天分别可额外领2、6金币。
每连续签到7天后重新累积签到天数（即重置签到天数：连续第8天签到时记为新的一轮签到的第一天，领1金币）

问题：计算每个用户2021年7月以来每月获得的金币数（该活动到10月底结束，11月1日开始的签到不再获得金币）。结果按月份、ID升序排序。

注：如果签到记录的in_time-进入时间和out_time-离开时间跨天了，也只记作in_time对应的日期签到了。

输出示例：

示例数据的输出结果如下：

uid	month	coin
101	202107	15
102	202110	7

解释：

101在活动期内连续签到了7天，因此获得1*7+2+6=15金币；

102在10.01~10.03连续签到3天获得5金币

10.04断签了，10.05~10.06连续签到2天获得2金币，共得到7金币。

1 数据准备

DROP TABLE IF EXISTS tb_user_log;
CREATE TABLE tb_user_log (
    uid INT COMMENT '用户ID',
    artical_id INT COMMENT '视频ID',
    in_time string COMMENT '进入时间',
    out_time string COMMENT '离开时间',
    sign_in int COMMENT '是否签到'
) ;

INSERT INTO tb_user_log(uid, artical_id, in_time, out_time, sign_in) VALUES
  (101, 0, '2021-07-07 10:00:00', '2021-07-07 10:00:09', 1),
  (101, 0, '2021-07-08 10:00:00', '2021-07-08 10:00:09', 1),
  (101, 0, '2021-07-09 10:00:00', '2021-07-09 10:00:42', 1),
  (101, 0, '2021-07-10 10:00:00', '2021-07-10 10:00:09', 1),
  (101, 0, '2021-07-11 23:59:55', '2021-07-11 23:59:59', 1),
  (101, 0, '2021-07-12 10:00:28', '2021-07-12 10:00:50', 1),
  (101, 0, '2021-07-13 10:00:28', '2021-07-13 10:00:50', 1),
  (101, 0, '2021-07-14 11:00:28', '2021-07-14 11:00:50', 1),
  (101, 0, '2021-07-15 11:59:28', '2021-07-16 00:01:20', 1),
  (102, 0, '2021-10-01 10:00:28', '2021-10-01 10:00:50', 1),
  (102, 0, '2021-10-02 10:00:01', '2021-10-02 10:01:50', 1),
  (102, 0, '2021-10-03 11:00:55', '2021-10-03 11:00:59', 1),
  (102, 0, '2021-10-04 11:00:45', '2021-10-04 11:00:55', 0),
  (102, 0, '2021-10-05 11:00:53', '2021-10-05 11:00:59', 1),
  (102, 0, '2021-10-06 11:00:45', '2021-10-06 11:00:55', 1);

2 问题分析

本题在牛客网上算比较难的一道题目，本地所要求的是每个用户每月获得的金币数：

领取金币规则：

（1）只要用户签到就能获取一枚金币
（2）连续签到第三天和第七天分别额外获取2枚和6金币
（3）连续签到7天后进行重置，按照规则（1）和（2）进行新一轮领币

分析：本题的维度是用户和月份，也就是最终结果是按照这两个来分组

第一步：根据领金币的规则，我们容易想到按照签到的连续性先进行分组，将连续的进行标记分成一组

第二步：在每一个连续的组里进行row_number()标记（rn），目的是获取连续的第三天和第七个天特殊位置。

第三步：根据第二步连续的天数中是否有超过7天的，然后以此进行分割，再进行分组

整体计算规则如下草图：

步骤三种，首先需要对为7天的地方进行标记（打标记的目的就是为了辅助计算）

标记的方法：rn % 7==0.如果为0则记为1，不为0则记为0

分组方法：按照滑动窗口求和的方法。此类分俩种一种是窗口变得越来越大的方法，另一种是窗口越来越小的方法。具体如下图所示

此两种算法对应的目标不一样：第一种窗口越来越大的方法是希望1标记位分组的时候能分到下面组中（标记位向下分组），第二种窗口越来越小的形式是希望标记为1能分到上一组数据中（或理解为进入到上一个数据桶中）（标记位向上）。具体整体分组如下图

步骤一：先判断连续性，做第一个分组标签

注意此处不要将是否签到值sign_in在where条件中过滤，此题连续真正的含义是：时间连续+签到的连续，如果sign_in在where条件中过滤则不能构成分组条件

select  uid
      ,in_time
      ,sign_in
      ,month
      -- 时间和签到同时的连续性
     , sum(if(diff < 1,1,0)) over(partition by uid,month,series_sign_in_flg order by in_time) as    series_flg

from(
select uid
      ,in_time
      ,sign_in
      ,month
    --求出当前值与上一个值的差值，如果为1具备连续的条件(时间的连续性)
      ,datediff(to_date(in_time),to_date(lag_in_time))  as diff
      --构造签到的连续性条件
      ,sum(if(sign_in!=lag_sign_in,1,0)) over(partition by uid,month order by in_time) as series_sign_in_flg
from
--求出上一行值
  (select uid
        ,in_time
        ,sign_in
        ,substr(in_time,1,7) as month
        ,lag(in_time,1,in_time) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_in_time
        ,lag(sign_in,1,sign_in) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_sign_in
  from tb_user_log
  where artical_id=0 and substr(in_time,1,7)>='2021-07' 
        and substr(in_time,1,7) <='2021-10'
  ) t
  ) t
where sign_in=1

+------+----------------------+----------+----------+-------------+
| uid  |       in_time        | sign_in  |  month   | series_flg  |
+------+----------------------+----------+----------+-------------+
| 101  | 2021-07-07 10:00:00  | 1        | 2021-07  | 1           |
| 101  | 2021-07-08 10:00:00  | 1        | 2021-07  | 1           |
| 101  | 2021-07-09 10:00:00  | 1        | 2021-07  | 1           |
| 101  | 2021-07-10 10:00:00  | 1        | 2021-07  | 1           |
| 101  | 2021-07-11 23:59:55  | 1        | 2021-07  | 1           |
| 101  | 2021-07-12 10:00:28  | 1        | 2021-07  | 1           |
| 101  | 2021-07-13 10:00:28  | 1        | 2021-07  | 1           |
| 101  | 2021-07-14 11:00:28  | 1        | 2021-07  | 1           |
| 101  | 2021-07-15 11:59:28  | 1        | 2021-07  | 1           |
| 102  | 2021-10-01 10:00:28  | 1        | 2021-10  | 1           |
| 102  | 2021-10-02 10:00:01  | 1        | 2021-10  | 1           |
| 102  | 2021-10-03 11:00:55  | 1        | 2021-10  | 1           |
| 102  | 2021-10-05 11:00:53  | 1        | 2021-10  | 0           |
| 102  | 2021-10-06 11:00:45  | 1        | 2021-10  | 0           |
+------+----------------------+----------+----------+-------------+

错误的分组标签生成：sign_in在where条件中过滤

select *
      -- 签到连续性标签
     , sum(if(datediff(to_date(in_time),to_date(lag_in_time)) < 1,1,0)) over(partition by uid,month order by in_time) as    series_flg
from
  (select uid
        ,in_time
        ,substr(in_time,1,7) as month
        ,lag(in_time,1,in_time) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_in_time
  from tb_user_log
  where artical_id=0 and substr(in_time,1,7)>='2021-07' 
        and substr(in_time,1,7) <='2021-10' and sign_in=1
  ) t

生成的结果如下:

步骤二：生成第二个标签，连续7天的分组标记

select *
      --生成连续7天的分组标记
      ,sum(case when rn % 7=0 then 1 else 0 end) over(partition by uid,month,series_flg order by in_time rows between current row and unbounded following) as series_7_flg
from 
(select *
--生成每组中的顺序标记
     , row_number() over(partition by uid,month,series_flg) as rn
from  
(select  uid
      ,in_time
      ,sign_in
      ,month
      -- 时间和签到同时的连续性
     , sum(if(diff < 1,1,0)) over(partition by uid,month,series_sign_in_flg order by in_time) as    series_flg

from(
select uid
      ,in_time
      ,sign_in
      ,month
    --求出当前值与上一个值的差值，如果为1具备连续的条件(时间的连续性)
      ,datediff(to_date(in_time),to_date(lag_in_time))  as diff
      --构造签到的连续性条件
      ,sum(if(sign_in!=lag_sign_in,1,0)) over(partition by uid,month order by in_time) as series_sign_in_flg
from
--求出上一行值
  (select uid
        ,in_time
        ,sign_in
        ,substr(in_time,1,7) as month
        ,lag(in_time,1,in_time) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_in_time
        ,lag(sign_in,1,sign_in) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_sign_in
  from tb_user_log
  where artical_id=0 and substr(in_time,1,7)>='2021-07' 
        and substr(in_time,1,7) <='2021-10'
  ) t
  ) t
where sign_in=1
) t
) t

对应结果如下：

步骤三：根据领取金币规则，在每个条件分组中求金币数

select uid
      ,month
      ,series_flg
      ,series_7_flg
      --按照领取金币规则求金币数
      ,case when cnt>=1 and cnt<3 then cnt
            when cnt>=3 and cnt<7 then cnt + 2
       else cnt + 2 + 6 end coin        
from
(select uid
      ,month
      ,series_flg
      ,series_7_flg
      ,count(1) as cnt
from
(select *
      --生成连续7天的分组标记
      ,sum(case when rn % 7=0 then 1 else 0 end) over(partition by uid,month,series_flg order by in_time rows between current row and unbounded following) as series_7_flg
from 
(select *
--生成每组中的顺序标记
     , row_number() over(partition by uid,month,series_flg) as rn
from  
(select  uid
      ,in_time
      ,sign_in
      ,month
      -- 时间和签到同时的连续性
     , sum(if(diff < 1,1,0)) over(partition by uid,month,series_sign_in_flg order by in_time) as    series_flg

from(
select uid
      ,in_time
      ,sign_in
      ,month
    --求出当前值与上一个值的差值，如果为1具备连续的条件(时间的连续性)
      ,datediff(to_date(in_time),to_date(lag_in_time))  as diff
      --构造签到的连续性条件
      ,sum(if(sign_in!=lag_sign_in,1,0)) over(partition by uid,month order by in_time) as series_sign_in_flg
from
--求出上一行值
  (select uid
        ,in_time
        ,sign_in
        ,substr(in_time,1,7) as month
        ,lag(in_time,1,in_time) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_in_time
        ,lag(sign_in,1,sign_in) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_sign_in
  from tb_user_log
  where artical_id=0 and substr(in_time,1,7)>='2021-07' 
        and substr(in_time,1,7) <='2021-10'
  ) t
  ) t
where sign_in=1
) t
) t
) t
group by uid,month,series_flg,series_7_flg
) t

+------+----------+-------------+---------------+-------+
| uid  |  month   | series_flg  | series_7_flg  | coin  |
+------+----------+-------------+---------------+-------+
| 102  | 2021-10  | 1           | 0             | 5     |
| 102  | 2021-10  | 0           | 0             | 2     |
| 101  | 2021-07  | 1           | 1             | 15    |
| 101  | 2021-07  | 1           | 0             | 2     |
+------+----------+-------------+---------------+-------+

分组1：series_flg表示连续签到分组。分组2：series_7_flg表示连续领取金币天数中有超过7天的，并以7天为分割点重新分组。

此处领取金币的算法为：1-2天为等于连续的天数，3-6天为连续的天数加2，当为7天时为连续天数加8

步骤四：获取用户每月的金币数，并按用户及月份升序排序

最终SQL如下

select uid,month,sum(coin) as coin
from(
select uid
      ,month
      ,series_flg
      ,series_7_flg
      --按照领取金币规则求金币数
      ,case when cnt>=1 and cnt<3 then cnt
            when cnt>=3 and cnt<7 then cnt + 2
       else cnt + 2 + 6 end coin        
from
(select uid
      ,month
      ,series_flg
      ,series_7_flg
      ,count(1) as cnt
from
(select *
      --生成连续7天的分组标记
      ,sum(case when rn % 7=0 then 1 else 0 end) over(partition by uid,month,series_flg order by in_time rows between current row and unbounded following) as series_7_flg
from 
(select *
--生成每组中的顺序标记
     , row_number() over(partition by uid,month,series_flg) as rn
from  
(select  uid
      ,in_time
      ,sign_in
      ,month
      -- 时间和签到同时的连续性
     , sum(if(diff < 1,1,0)) over(partition by uid,month,series_sign_in_flg order by in_time) as    series_flg

from(
select uid
      ,in_time
      ,sign_in
      ,month
    --求出当前值与上一个值的差值，如果为1具备连续的条件(时间的连续性)
      ,datediff(to_date(in_time),to_date(lag_in_time))  as diff
      --构造签到的连续性条件
      ,sum(if(sign_in!=lag_sign_in,1,0)) over(partition by uid,month order by in_time) as series_sign_in_flg
from
--求出上一行值
  (select uid
        ,in_time
        ,sign_in
        ,substr(in_time,1,7) as month
        ,lag(in_time,1,in_time) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_in_time
        ,lag(sign_in,1,sign_in) over(partition by uid,substr(in_time,1,7) order by in_time) as lag_sign_in
  from tb_user_log
  where artical_id=0 and substr(in_time,1,7)>='2021-07' 
        and substr(in_time,1,7) <='2021-10'
  ) t
  ) t
where sign_in=1
) t
) t
) t
group by uid,month,series_flg,series_7_flg
) t
)t
group by uid,month 
order by uid,month

最终结果如下：

+------+----------+-------+
| uid  |  month   | coin  |
+------+----------+-------+
| 101  | 2021-07  | 17    |
| 102  | 2021-10  | 7     |
+------+----------+-------+

注意此处与所给的答案有出入是因为造的数据中增加了2条数据，便于测试。

3 小结

本文给出了一种连续签到领取金币的一种通用解法，通过窗口函数生成标签值进行辅助计算。通过本题可以收获以下知识点：

（1）连续签到的判断方法（注意时间连续+签到的连续）
（2）断点（按照某种条件）重分组的方法（如本题的连续7天后为一个断点）

         1) 断点同上：order by in_time rows between current row and unbounded   

                     following（上边界滑动，下边界固定）

         2）断点同下： order by in_time（下边界滑动，上边界固定）

（3）周期的计算方法，如本题7天一个循环（采用求余的方法：rn%7==0）