Sql server SQL Server 2012（T-SQL）按组选择日到月

你好

我将每日数据汇总为每月数据

TABLE daily_data:
item_from,
item_to,
agg_day,
average_day_rate,
min_day_rate,
max_day_rate,
closing_day_rate

TABLE monthly_data:
item_from,
item_to,
agg_month,
average_month_rate,
min_month_rate,
max_month_rate,
closing_month_rate

选择数据：

DECLARE @rundate       DATE;
DECLARE @yesterday     DATE;
DECLARE @firstofmonth  DATE;

SET @runday        = CONVERT(DATE, GETDATE())
SET @yesterday     = CONVERT(DATE, DATEADD(dd,-1,GETDATE()))
SET @firstofmonth  = DATEADD(dd, 1, EOMONTH(@yesterday, -1))

SELECT
  item_from,
  item_to,
  DATEADD(dd, 1, EOMONTH(agg_day, -1))                                     agg_month,
  SUM(average_day_rate)/COUNT(DISTINCT(agg_day))                          average_month_rate,
  MIN(min_day_rate)                                                       min_month_rate,
  MAX(max_day_rate)                                                       max_month_rate,
  (SELECT closing_day_rate FROM daily_data WHERE agg_day = MAX(agg_day))  closing_month_rate
FROM
  daily_data
WHERE
  agg_day >= @firstofmonth
  and agg_day < @runday
GROUP BY
  item_from,
  item_to,
  DATEADD(dd, 1, EOMONTH(agg_day, -1))

结果: 聚合不能出现在WHERE子句中，除非它位于HAVING子句或select列表中包含的子查询中，并且被聚合的列是外部引用

我试着在小组中使用aggday，将where改为having，但没有成功。 在PL-SQL中，我将使用

SELECT
  item_from,
  item_to,
  trunc(agg_day, MONTH)                                                  agg_month,
  SUM(average_day_rate)/COUNT(DISTINCT(agg_day))                          average_month_rate,
  MIN(min_day_rate)                                                       min_month_rate,
  MAX(max_day_rate)                                                       max_month_rate,
  (SELECT closing_day_rate FROM daily_data WHERE agg_day = MAX(agg_day))  closing_month_rate
FROM
  daily_data
WHERE
  agg_day >= @firstofmonth
  and agg_day < @runday
GROUP BY
  item_from,
  item_to,
  trunc(agg_day, MONTH)

请提供帮助。

解决方案：

DECLARE @rundate        DATE;
DECLARE @yesterday      DATE;
DECLARE @firstofmonth   DATE;
DECLARE @lastdayclosing SMALLMONEY;

SET @runday        = CONVERT(DATE, GETDATE())
SET @yesterday     = CONVERT(DATE, DATEADD(dd,-1,GETDATE()))
SET @firstofmonth  = DATEADD(dd, 1, EOMONTH(@yesterday, -1))
SET @lastdayclosing = (SELECT close_rate FROM exchange_rate_daily WHERE exr_date = @yesterday)

SELECT
  item_from,
  item_to,
  @firstofmonth                                   agg_month,
  SUM(average_day_rate)/COUNT(DISTINCT(agg_day))  average_month_rate,
  MIN(min_day_rate)                               min_month_rate,
  MAX(max_day_rate)                               max_month_rate,
  @lstdayclosing                                  closing_month_rate
FROM
  daily_data
WHERE
  agg_day >= @firstofmonth
  and agg_day < @runday
GROUP BY
  item_from,
  item_to
;

解决方案：

DECLARE @rundate        DATE;
DECLARE @yesterday      DATE;
DECLARE @firstofmonth   DATE;
DECLARE @lastdayclosing SMALLMONEY;

SET @runday        = CONVERT(DATE, GETDATE())
SET @yesterday     = CONVERT(DATE, DATEADD(dd,-1,GETDATE()))
SET @firstofmonth  = DATEADD(dd, 1, EOMONTH(@yesterday, -1))
SET @lastdayclosing = (SELECT close_rate FROM exchange_rate_daily WHERE exr_date = @yesterday)

SELECT
  item_from,
  item_to,
  @firstofmonth                                   agg_month,
  SUM(average_day_rate)/COUNT(DISTINCT(agg_day))  average_month_rate,
  MIN(min_day_rate)                               min_month_rate,
  MAX(max_day_rate)                               max_month_rate,
  @lstdayclosing                                  closing_month_rate
FROM
  daily_data
WHERE
  agg_day >= @firstofmonth
  and agg_day < @runday
GROUP BY
  item_from,
  item_to
;

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为什么不将agg_day=MAXagg_day的每日数据中的选择收盘日利率放入cte？或者更好：将结果直接放入变量？为什么不将agg_day=MAXagg_day的每日数据中的选择收盘日利率放入cte？或者更好：将结果直接放入变量？