编辑： begin_date 和 end_date 是任何表中的类型 DATE 列。
我有以下维度表，它提供了从 1980 年到 2500 年每个月的总天数：
CREATE TABLE total_days
(
from_date DATE,
to_date DATE,
days_in_month SMALLINT
);
from_date to_date days_in_month
1980-01-01 1980-01-31 31
1980-02-01 1980-02-29 29
...
2500-11-01 2500-11-30 30
2500-12-01 2500-12-31 31
如果要向 begin_date 添加 360 个月，我应该如何构造 SQL 查询以获得准确的 end_date？.. 我是否需要以任何方式更改维度表以实现我的目标？
编辑：必须在不使用任何本机 SQL 日期算术函数的情况下执行日期算术。必须通过在维度表中查找 begin_date 来完成。
我想你有你的理由 - 这是一个非常简单的技巧 - 假设事实表每个月都有一行 - 添加一个代表月份编号的新列，从 1 开始并按时间顺序自动递增，而不是重新开始每年。
SELECT B.*
FROM SO_total_days2 A
INNER JOIN SO_total_days2 B ON B.monthnumber = A.monthnumber + 360
WHERE A.from_date = '2010-01-01'
from_date to_date days_in_month monthnumber
1980-01-01 1980-01-31 31 1
1980-02-01 1980-02-29 29 2
1980-03-03 1980-03-31 31 3
...
1981-01-01 1981-01-31 31 13
1981-12-01 1981-12-31 31 24
...
1985-01-01 1985-01-31 31 49
1985-12-01 1985-12-31 31 60
这就是我想象你的“事实表”的样子：
declare @dt datetime
set @dt = '7-1-2012'
;
with date_table as (
select @dt as [Start Date],
dateadd(d,-1,dateadd(mm,1,@dt)) as [End Date],
datepart(d,dateadd(d,-1,dateadd(mm,1,@dt))) as [Days]
union ALL
select dateadd(mm, 1, [Start Date]) as [Start Date],
dateadd(d,-1,dateadd(mm,1,dateadd(mm, 1, [Start Date]))) as [End Date],
datepart(d,dateadd(d,-1,dateadd(mm,1,dateadd(mm, 1, [Start Date])))) as [Days]
from date_table
where dateadd(mm, 1, [Start Date]) <= dateadd(m,500,@dt))
select [Start Date], [End Date], [Days]
into #temp
from date_table
option (MAXRECURSION 0)
这是选择日期。（注意这些语句中没有包含 DATEADD 或 DATEPART）
select finish.[Start Date], finish.[End Date], finish.[Days]
from (select rownum
from (select [Start Date], [End Date], [Days], row_number() over (order by [Start Date]) as rownum
from #temp) as x
where x.[Start Date] = '2012-07-01 00:00:00.000' ) as start
join (select [Start Date], [End Date], [Days],
row_number() over (order by [Start Date]) as rownum
from #temp) as finish
on finish.rownum = start.rownum + 360
我在下面阅读了您的评论...如果您想总结日子或其他事情，这就是您可以做到的：（因此从 2012 年 7 月 1 日开始，持续 360 个月... date_diff_days 结果将是360 个月的总天数...使用我制作的#temp 表...我假设它与您的事实表相似...我有 10957 天）
select sum(dayscount.[Days]) as date_diff_days
from (select rownum
from (select [Start Date], [End Date], [Days], row_number() over (order by [Start Date]) as rownum
from #temp) as x
where x.[Start Date] = '2012-07-01 00:00:00.000' ) as start
join (select [Start Date], [End Date], [Days],
row_number() over (order by [Start Date]) as rownum
from #temp) as finish
on finish.rownum = start.rownum + 360
join (select [Start Date], [End Date], [Days],
row_number() over (order by [Start Date]) as rownum
from #temp) as dayscount
on dayscount.rownum >= start.rownum and
dayscount.rownum < finish.rownum
如果我要与数据库无关，我会稍微更改一下事实表：
CREATE TABLE total_days
(
year INT,
month TINYINT,
from_date DATE,
to_date DATE,
days_in_month SMALLINT
);
year month from_date to_date days_in_month
------------------------------------------------
1980 1 1980-01-01 1980-01-31 31
1980 2 1980-02-01 1980-02-29 29
...
2500 11 2500-11-01 2500-11-30 30
2500 12 2500-12-01 2500-12-31 31
然后你可以使用类似的东西：
SELECT td.*
FROM
total_days AS td
CROSS JOIN
( SELECT year, month
FROM total_days
WHERE from_date <= @StartingDate
AND @StartingDate <= to_date
) AS st
CROSS JOIN
( SELECT 360 AS add_months ) AS param
WHERE td.year = st.year + ( st.month -1 + add_months ) / 12
AND td.month = 1 + ( st.month - 1 + add_months ) % 12 )
;
或更简单的（但更难优化效率：
WHERE 12 * td.year + td.month =
12 * st.year + st.month + add_months
为什么是事实表？大多数数据库对日期/时间操作都有本机支持。在 MS SQL Server 中，您可以使用DATEADD执行此操作。
我看到您用“informix”标记了您的问题，但没有在您的问题中指定任何版本详细信息。 这是来自 IBM Informix 11.50 的 ADD_MONTHS 函数。
你可以使用间隔
mysql> SELECT '2008-12-31 23:59:59' + INTERVAL 1 month;
+------------------------------------------+
| '2008-12-31 23:59:59' + INTERVAL 1 month |
+------------------------------------------+
| 2009-01-31 23:59:59 |
+------------------------------------------+
1 row in set (0.00 sec)
mysql> select now();
+---------------------+
| now() |
+---------------------+
| 2012-07-03 12:27:46 |
+---------------------+
1 row in set (0.00 sec)
mysql> SELECT now() + INTERVAL 30 month;
+---------------------------+
| now() + INTERVAL 30 month |
+---------------------------+
| 2015-01-03 12:27:49 |
+---------------------------+
1 row in set (0.00 sec)
编辑：
mysql> SELECT STR_TO_DATE('01,5,2013','%d,%m,%Y') + interval 30 month;
+---------------------------------------------------------+
| STR_TO_DATE('01,5,2013','%d,%m,%Y') + interval 30 month |
+---------------------------------------------------------+
| 2015-11-01 |
+---------------------------------------------------------+
1 row in set (0.00 sec)
编辑2：
mysql> show create table tiempo;
+--------+------------------------------------------------------------------------------------------------+
| Table | Create Table |
+--------+------------------------------------------------------------------------------------------------+
| tiempo | CREATE TABLE `tiempo` (
`fecha` datetime DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1 |
+--------+------------------------------------------------------------------------------------------------+
1 row in set (0.00 sec)
mysql> select fecha + interval 20 month from tiempo;
+---------------------------+
| fecha + interval 20 month |
+---------------------------+
| NULL |
| 2001-10-02 02:02:02 |
+---------------------------+
2 rows in set (0.00 sec)
如果您的起始日期与列出的完全相同，并且您总是添加月份，您可以使用这个：
SELECT max (to_date)
FROM (SELECT ROW_NUMBER () OVER (ORDER BY from_date) AS Row,
from_date,
to_date,
days_in_month
FROM total_days
WHERE from_date > '1/1/1982'
GROUP BY from_date, to_date, days_in_month) MyDates
WHERE Row <= 360