在获取每个日期的最近值的同时获取每个日期的值总和？

Question

我有一个表格，动作，显示用户使用他们的卡。 是这样的：

movementdate	expense	deposit	balance
2021-07-09 19:23:49	0	2	2
2021-07-09 20:40:13	0	6	8
2021-07-09 13:50:01	0	2	10
2021-07-10 11:50:31	5	0	5
2021-07-10 12:59:15	0	5	10
2021-07-10 18:01:39	10	0	0

它显示了他们的存款和费用，还显示了他们的最终余额。 我有这个查询，可以获取每天的存款和费用总额；

select EXTRACT(day FROM m.movementdate) as d, EXTRACT(month FROM m.movementdate) as m, 
       EXTRACT(year FROM m.movementdate) as y, sum(m.expense) as TotalExpense, 
       sum (m.deposit) as TotalDeposit
from movements m 
group by 3,2,1

但我还想获得他们每天余额的最后记录。

类似这样的：

Date	TotalExpense	TotalDeposit	Balance
2021-07-09	0	10	10
2021-07-10	15	5	0

我怎样才能得到最后一部分？我找到了一些方法来获取当天的最后一个余额，但我不知道如何将它包含在同一个查询中。

Answer 1

另一种达到预期效果的方法是使用窗口函数row_number()

这里是db<>fiddle

with cte as
(
  select 
    movementdate::date as dt,
    EXTRACT(day FROM movementdate) as d,
    EXTRACT(month FROM movementdate) as m, 
    EXTRACT(year FROM movementdate) as y,
    sum(expense) over (partition by movementdate::date) as expense,
    sum(deposit) over (partition by movementdate::date) as deposit,
    balance,
    row_number() over (partition by movementdate::date order by movementdate desc) as rn
  from movements
)

select 
  dt,
  d, 
  m, 
  y, 
  expense, 
  deposit, 
  balance
from cte
where rn = 1
order by
  1, 2, 3

输出：

*-------------------------------------------------------*
|  dt          d    m   y     expense   deposit  balance|
*-------------------------------------------------------*
| 2021-07-09    9   7   2021    0         10       8    |
| 2021-07-10    10  7   2021    15         5       0    |
*-------------------------------------------------------*

Answer 2

一种方法使用数组：

select EXTRACT(day FROM m.movementdate) as d,
       EXTRACT(month FROM m.movementdate) as m, 
       EXTRACT(year FROM m.movementdate) as y,
       sum(m.expense) as TotalExpense, 
       sum (m.deposit) as TotalDeposit,
       (array_agg(m.balance order by m.movementdate desc))[1] as balance
from movements m 
group by 3,2,1;

我还建议您使用m.movementdate::date 而不是定义前三列。

Answer 3

另一种选择是使用PostgreSQL LATERAL Subquery

类似：

WITH movements_per_day AS (
  select m.movementdate::date AS dt
     , EXTRACT(day FROM m.movementdate) as d
     , EXTRACT(month FROM m.movementdate) as m
     , EXTRACT(year FROM m.movementdate) as y
     , sum(m.expense) as TotalExpense
     , sum (m.deposit) as TotalDeposit
  from movements m
  group by 1,2,3,4
)
SELECT movements_per_day.*, last_movement.balance
  FROM movements_per_day
  CROSS JOIN LATERAL (
    SELECT movements.balance
    FROM movements
    WHERE movements.movement_date::date = movements_per_day.dt
    ORDER BY movements.movementdate DESC 
    LIMIT 1
  ) AS last_movement

正如其他答案所暗示的，使用m.movementdate::date 而不是前三列会更简单一些。