SQL 计算一段时间内的累计值

Question

我正在尝试计算自 2020 年 1 月 1 日以来随时间推移的累计收入。 我有以下架构的用户级收入数据

create table revenue
(
  game_id        varchar(255),
  user_id        varchar(255),
  amount         int,
  activity_date  varchar(255)
);

insert into revenue
  (game_id, user_id, amount, activity_date)
values
  ('Racing', 'ABC123', 5, '2020-01-01'),
  ('Racing', 'ABC123', 1, '2020-01-04'),
  ('Racing', 'CDE123', 1, '2020-01-04'),
  ('DH', 'CDE123', 100, '2020-01-03'),
  ('DH', 'CDE456', 10, '2020-01-02'),
  ('DH', 'CDE789', 5, '2020-01-02'),
  ('DH', 'CDE456', 1, '2020-01-03'),
  ('DH', 'CDE456', 1, '2020-01-03');

预计 Output

Game    Age    Cum_rev    Total_unique_payers_per_game
Racing  0      5          2
Racing  1      5          2
Racing  2      5          2
Racing  3      7          2
DH      0      0          3
DH      1      15         3
DH      2      117        3
DH      3      117        3

年龄计算为交易日期与 2020-01-01 之间的差异。 我正在使用以下逻辑

SELECT game_id, DATEDIFF(activity_date ,'2020-01-01') as Age,count(user_id) as Total_unique_payers
from REVENUE

SQL fiddle如何计算累计收益？

Answer 1

对于以下内容，您需要一个支持over()子句（MySQL 8+）的 MySQL 版本 - 我使用了 MariaDB 10.4 下面（当我尝试时，MySQL 8 无法在该站点工作）

 create table revenue ( game_id varchar(255), user_id varchar(255), amount int, activity_date varchar(255) ); insert into revenue (game_id, user_id, amount, activity_date) values ('Racing', 'ABC123', 5, '2020-01-01'), ('Racing', 'ABC123', 1, '2020-01-04'), ('Racing', 'CDE123', 1, '2020-01-04'), ('DH', 'CDE123', 100, '2020-01-03'), ('DH', 'CDE456', 10, '2020-01-02'), ('DH', 'CDE789', 5, '2020-01-02'), ('DH', 'CDE456', 1, '2020-01-03'), ('DH', 'CDE456', 1, '2020-01-03');

  ✓ ✓

 SELECT game_id, user_id, activity_date, amount, sum(amount) over(order by activity_date, user_id) as running_sum, (select count(distinct user_id) from revenue) as Total_unique_payers from revenue order by activity_date, user_id

 游戏ID | 用户 ID | 活动日期 | 金额 | 运行总和 |  Total_unique_payers:------ |:-------- |:------------ |  -----: |  ----------: |  ------------------: 赛车 |  ABC123 |  2020-01-01 |  5 |  5 |  4 DH |  CDE456 |  2020-01-02 |  10 |  15 |  4 DH |  CDE789 |  2020-01-02 |  5 |  20 |  4 DH |  CDE123 |  2020-01-03 |  100 |  120 |  4 DH |  CDE456 |  2020-01-03 |  1 |  122 |  4 DH |  CDE456 |  2020-01-03 |  1 |  122 |  4 赛车 |  ABC123 |  2020-01-04 |  1 |  123 |  4 赛车 |  CDE123 |  2020-01-04 |  1 |  124 |  4

db<> 在这里摆弄

更改 over 子句中的计算顺序会影响运行总和的计算方式：例如

SELECT game_id, user_id, activity_date, amount, sum(amount) over(order by game_id DESC, activity_date, user_id) as running_sum, (select count(distinct user_id) from revenue) as Total_unique_payers from revenue order by game_id DESC, activity_date, user_id

 游戏ID | 用户 ID | 活动日期 | 金额 | 运行总和 |  Total_unique_payers:------ |:-------- |:------------ |  -----: |  ----------: |  ------------------: 赛车 |  ABC123 |  2020-01-01 |  5 |  5 |  4 赛车 |  ABC123 |  2020-01-04 |  1 |  6 |  4 赛车 |  CDE123 |  2020-01-04 |  1 |  7 |  4 DH |  CDE456 |  2020-01-02 |  10 |  17 |  4 DH |  CDE789 |  2020-01-02 |  5 |  22 |  4 DH |  CDE123 |  2020-01-03 |  100 |  122 |  4 DH |  CDE456 |  2020-01-03 |  1 |  124 |  4 DH |  CDE456 |  2020-01-03 |  1 |  124 |  4

db<> 在这里摆弄

Answer 2

使用 MySQL 5.7 的唯一方法是使用它的变量系统，尽管它有效。 它模拟了@Used_By_Already 在他的回答中使用的窗口函数

既然你提到你关心差距，你首先需要创建一个日期表，这很容易做到：

create table dates_view (
  date_day date
);

insert into dates_view
select date_add( '2019-12-31', INTERVAL @rownum:=@rownum+1 day ) as date_day
from (
   select 0 union select 1 union select 2 union select 3 
   union select 4 union select 5 union select 6 
   union select 7 union select 8 union select 9
) a, (
   select 0 union select 1 union select 2 union select 3 
   union select 4 union select 5 union select 6 
   union select 7 union select 8 union select 9
) b, (select @rownum:=0) r;

-- Note: each set of select union above will multiply the number 
-- of days by 10, so if you need more days in your table just add more
-- set as above "a" or "b" sets

在拥有 Dates 表后，您必须将其与当前的revenue表交叉连接，问题是您希望玩家数量独立于累积amount ，因此您需要在子查询中独立计算它。

您还需要计算您的revenue表的max(activity_date) ，以便将结果限制在它之前。

所以下面的查询将做到这一点（基于您当前的示例数据）：

set @_sum:=0;       -- Note: this two lines depends on the client
set @_currGame:=''; -- you are using. Some accumulate variable per session
                    -- some doesn't, below site, for instance does

select a.game_id,
       a.age,
       case when @_currGame = game_id 
            then @_sum:=coalesce(samount,0) + @_sum
            else @_sum:=coalesce(samount,0) end as Cum_rev,
       a.Total_unique_payers_per_game,
       @_currGame := game_id varComputeCurrGame
from 
    (
    select players.game_id, 
           rev.samount,
           datediff(dv.date_day, '2020-01-01') age,
           players.noPlayers Total_unique_payers_per_game
       from (select @_sum:=0) am,
            dates_view dv
             cross join (select max(activity_date) maxDate from revenue) md 
               on dv.date_day <= md.maxDate
             cross join (select game_id, count(distinct user_id) noPlayers 
                           from revenue group by game_id) players
             left join (select game_id, activity_date, sum(amount) samount 
                          from revenue group by game_id, activity_date) rev
                on players.game_id = rev.game_id
                   and dv.date_day = rev.activity_date
    ) a,
    (select @_sum:=0) s,
    (select @_currGame='') x
order by a.game_id desc, a.age;

这将导致：

  game_id   age  Cum_rev  Total_unique_payers_per_game   varComputeCurrGame
   Racing    0      5             2                            Racing
   Racing    1      5             2                            Racing
   Racing    2      5             2                            Racing
   Racing    3      7             2                            Racing
   DH        0      0             3                            DH    
   DH        1      15            3                            DH    
   DH        2      117           3                            DH    
   DH        3      117           3                            DH  

看到它在这里工作（你需要运行它）： https://www.db-fiddle.com/f/qifZ6hmpvcSZYwhLDv613d/2

这是支持窗口功能的 MySQL 8.x 版本：

select distinct agetable.game_id,
       agetable.age,
       sum(coalesce(r1.amount,0)) 
             over (partition by agetable.game_id 
                     order by agetable.game_id, agetable.age) as sm,
       agetable.ttplayers
from
    (
    select r.game_id, dv.date_day, datediff(dv.date_day, '2020-01-01') age, p.ttplayers
    from dates_view dv
          cross join (select distinct game_id, activity_date from revenue) r 
            on dv.date_day <= (select max(activity_date) from revenue)
          left join (select game_id, count(distinct user_id) ttplayers from revenue group by game_id) p
            on r.game_id = p.game_id
    group by r.game_id desc, dv.date_day, age, p.ttplayers
    ) agetable
    left join revenue r1
      on agetable.date_day = r1.activity_date
         and r1.game_id = agetable.game_id
order by agetable.game_id desc, agetable.age