Question

我有一张桌子，我正在查看授权及其每月价值。我已经将auth的总值取了几个月，然后除以auth有效的月数给我每月的金额。为了聚合数据，我需要以某种方式在该身份验证时间框架内为每个月分配月度值。例如，我有一个auth，开始于2017年8月1日，结束于1月31日，持续6个月。 auth的总价值为3559.50美元，因此每月为593.25美元。所以我需要以某种方式在8月到1月期间每个月分配593.25美元。我在结果中寻找这个。

Month    member    amount
8/1/2017    12345   593.25
9/1/2017    12345   593.25
10/1/2017    12345   593.25
11/1/2017    12345   593.25
12/1/2017    12345   593.25
1/1/2018    12345   593.25

以下是示例数据：

 create table #temp
 (month date,
 memberId varchar(5),
 auth_datefrom date,
 auth_dateto date,
 authmonths int,
 est_monthly_payment money)

INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','12345','8/1/2017','9/30/2017','2','762.75');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','67890','8/1/2017','9/30/2017','2','2440.8');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','23456','8/1/2017','6/30/2018','11','443.78');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','34567','8/1/2017','8/31/2017','1','1708.56');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','45678','8/1/2017','8/31/2017','1','4881.6');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','56789','8/1/2017','11/1/2017','3','996.66');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','67890','8/1/2017','6/30/2018','11','443.78');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','78901','8/1/2017','8/31/2017','1','1708.56');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','89012','8/1/2017','8/31/2017','1','4881.6');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','90123','8/1/2017','2/28/2018','7','813.6');
INSERT INTO #temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','1234','8/1/2017','6/30/2018','11','443.78');

Answer 1

这是一个猜测，但假设startdate是固定的，就像在示例代码中一样，这可能会有效。

CREATE TABLE #DateTable (Dates DATETIME, ID INT IDENTITY )

DECLARE @Counter INT
SET @Counter = 0

WHILE @Counter < 481

BEGIN 
INSERT INTO #DateTable
SELECT DATEADD (MONTH, @Counter, '2017-08-01') 
SET @Counter = @Counter + 1

END

SELECT D.*, T.member_Id, T.auth_datefrom, T.auth_dateto, T.authmonths, 
T.est_monthly_payment 
FROM #DateTable AS D
CROSS JOIN  #temp AS T 
WHERE D.ID <= T.authmonths

Answer 2

所以我所做的只是将其转换为单个用户。您可以将其转换为proc，然后为您提供结果集，并且可以为每行auths调用proc并添加到结果集。如果你想在一次操作中完成它，你需要做一些可以获得日期的CTE。

本质上，CTE循环遍历行的日期范围，并在每个月拆分，然后您只需将总auth除以之间的持续时间。

我是这样做的。

declare @temp table
 ([month] date,
 member_id varchar(5),
 auth_datefrom date,
 auth_dateto date,
 authmonths int,
 est_monthly_payment money);

INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','12345','8/1/2017','9/30/2017','2','762.75');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','67890','8/1/2017','9/30/2017','2','2440.8');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','23456','8/1/2017','6/30/2018','11','443.78');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','34567','8/1/2017','8/31/2017','1','1708.56');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','45678','8/1/2017','8/31/2017','1','4881.6');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','56789','8/1/2017','11/1/2017','3','996.66');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','67890','8/1/2017','6/30/2018','11','443.78');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','78901','8/1/2017','8/31/2017','1','1708.56');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','89012','8/1/2017','8/31/2017','1','4881.6');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','90123','8/1/2017','2/28/2018','7','813.6');
INSERT INTO @temp (Month,member_id,auth_datefrom,auth_dateto,authmonths,Est_Monthly_Payment) VALUES ('8/1/2017','1234','8/1/2017','6/30/2018','11','443.78');

select * from @temp
WHERE
    member_id = '12345';

declare @startDate DateTime, @endDate DateTime;
select @startDate = auth_datefrom, @endDate = auth_dateto from @temp where member_id = '12345';

; with GetDates AS
(
    select dateadd(month, 1, @startDate) as TheDate
    UNION ALL
    select dateadd(month, 1, TheDate) from GetDates
    where TheDate < @endDate
)
select 
    t.member_id, gd.TheDate, (t.est_monthly_payment / t.authmonths) as monthly_payment

from 
    GetDates gd
    cross join @temp t

where
    t.member_id = '12345'
;

将值分配给连续的月份

2 个答案: