在 SQL 上计算滚动平均值

Question

我需要编写一个查询，给出 1 月 31 日每天处理的总交易金额的滚动 3 天平均值

我使用了下面的查询，但我在检索信息时遇到问题，因为它不断给出此错误：

查询错误：错误：列“sum_pday.sumamount”必须出现在 GROUP BY 子句中或在聚合函数中使用

select 
    Sum_pDay.day
    , Sum_pDay.Month
    , Sum_pDay.Year
     ,  avg (Sum_pDay.SumAmount) over (order by Sum_pDay.day rows between 2 preceding and current row) as Avg
from (
  select 
    extract (year from transaction_time) as Year
      , extract (month from transaction_time) as Month
      , extract (day from transaction_time) as day
      , sum (transaction_amount) as SumAmount
     from
        transactions
     group by Year, Month, day
     order by 3) as Sum_pDay
group by 1,2,3

Answer 1

错误消息告诉您需要按该字段进行分组，因此这可能是解决方案：

select 
    Sum_pDay.day
    , Sum_pDay.Month
    , Sum_pDay.Year
     ,  avg (Sum_pDay.SumAmount) over (order by Sum_pDay.day rows between 2 preceding and current row) as Avg
from (
  select 
    extract (year from transaction_time) as Year
      , extract (month from transaction_time) as Month
      , extract (day from transaction_time) as day
      , sum (transaction_amount) as SumAmount
     from
        transactions
     group by Year, Month, day
     order by 3) as Sum_pDay
group by 1,2,3, 4

如果有效请告诉我。

Answer 2

当您使用

GROUP BY

时，错误消息很清楚，每列必须位于分组依据中或具有聚合函数。

在您的情况下，您可以对组中已有的金额进行求和

更好的是变体 2，因为你真的不需要聚合

CREATE TABLe transactions (transaction_time date,transaction_amount decimal(10,2))

CREATE TABLE

select 
    Sum_pDay.day
    , Sum_pDay.Month
    , Sum_pDay.Year
     ,  avg (SUM(Sum_pDay.SumAmount)) over (order by Sum_pDay.day rows between 2 preceding and current row) as Avg
from (
  select 
    extract (year from transaction_time) as Year
      , extract (month from transaction_time) as Month
      , extract (day from transaction_time) as day
      , sum (transaction_amount) as SumAmount
     from
        transactions
     group by Year, Month, day
     order by 3) as Sum_pDay
group by 1,2,3

日	月	年	平均

SELECT 0

select 
    Sum_pDay.day
    , Sum_pDay.Month
    , Sum_pDay.Year
     ,  avg (Sum_pDay.SumAmount) over (order by Sum_pDay.day rows between 2 preceding and current row) as Avg
from (
  select 
    extract (year from transaction_time) as Year
      , extract (month from transaction_time) as Month
      , extract (day from transaction_time) as day
      , sum (transaction_amount) as SumAmount
     from
        transactions
     group by Year, Month, day
     order by 3) as Sum_pDay

日	月	年	平均

SELECT 0

小提琴