Database 比较日期，计算它们之间的差异，博士后

我为每个用户设置了一个日期列和一个余额列。每次用户进行事务处理时，都会在此表中添加一个新行。可能是用户在一天内进行了15次交易，而在5天内根本没有交易

像这个

  date                balance
2017-06-01              95.63
2017-06-01              97.13
2017-06-01              72.14
2017-06-06              45.04
2017-06-08              20.04
2017-06-09              10.63
2017-06-09              -29.37
2017-06-09              -51.35
2017-06-13              -107.55
2017-06-13              -101.35
2017-06-15              -157.55
2017-06-16              -159.55
2017-06-17              -161.55

目标是选择当天交易中的正、负交易，计算其平均值或最小值，并将其视为一笔交易。如果第二天没有交易，则应使用前一天的金额。 这意味着一个月内的每一天我都应该计算利息，如果余额没有更新，那么应该使用前一天的余额。 假设我的桌子看起来像

date    balance
1/6/2017    72.14
6/2/2017    72.14
6/3/2017    72.14
6/4/2017    72.14
6/5/2017    72.14
6/6/2017    45.04
7/6/2017    45.04
8/6/2017    20.04
9/6/2017    -51.35
10/6/2017   -51.35
11/6/2017   -51.35
12/6/2017   -51.35
13/06/2017  -107.55
14/06/2017  -107.55
15/06/2017  -157.55
16/06/2017  -159.55
17/06/2017  -161.55

我添加了缺失的那些日子，并将重复的日子分组

完成此操作后，我可以选择正余额天数，例如8天，计算平均正余额，并将其乘以0.4%

8*58.8525*0.004=0.23

负平衡也应该如此。但负余额天数的利率不同，例如9乘以该天数内的平均负余额和8.49%

9*-99.90555556*0.00849=-0.848

因此，我的预期结果是只有这两列

 Neg        Pos
-0.848     0.23

我怎样才能在博士后做到这一点？函数重叠实际上没有帮助，因为我需要指定日期

此外，我也不知道该怎么做

循环日期，查看是否有重复项

查看缺失的天数，并使用这些缺失天数的上一个余额

---

请试试这个。。用表名替换表

with cte as
(
   Select "date" as date
         ,min(balance) as balance
         ,lead("date")  over(order by "date") next_date
         ,Coalesce(ABS("date" - lead("date")  over(order by "date")),1) date_diff 
from table
   group by "date"
),
cte2 as
(
   Select date_diff*balance as tot_bal , date_diff
   from cte 
   Where balance > 0  
),
cte3 as
(
   Select date_diff*balance as tot_bal , date_diff
   from cte 
   Where balance < 0  
)

Select (sum(cte2.tot_bal) / sum(cte2.date_diff) ) * 0.004 as pos
,(sum(cte3.tot_bal) / sum(cte3.date_diff) ) * 0.00849 as neg
from cte2
     ,cte3;

---

请试试这个。。用表名替换表

with cte as
(
   Select "date" as date
         ,min(balance) as balance
         ,lead("date")  over(order by "date") next_date
         ,Coalesce(ABS("date" - lead("date")  over(order by "date")),1) date_diff 
from table
   group by "date"
),
cte2 as
(
   Select date_diff*balance as tot_bal , date_diff
   from cte 
   Where balance > 0  
),
cte3 as
(
   Select date_diff*balance as tot_bal , date_diff
   from cte 
   Where balance < 0  
)

Select (sum(cte2.tot_bal) / sum(cte2.date_diff) ) * 0.004 as pos
,(sum(cte3.tot_bal) / sum(cte3.date_diff) ) * 0.00849 as neg
from cte2
     ,cte3;