我如何return 60组月份和年份数据在一个月份和年份列中？

Question

我得到了两个整数类型的变量，@month 和 @year（因此月份是 1、2、3、... 9、10、11、12，年份是四个-数字年份）。

我想要做的是创建一个 table 值函数，它将接受这两个参数和 return 过去 60 个月，然后到 2005 年结束。

因此，例如，对于 @month = 1 和 @year = 2018，我希望 return 这个 table:

+----+------+
| 1  | 2018 |
+----+------+
| 12 | 2017 |
+----+------+
| 11 | 2017 |
+----+------+
| 10 | 2017 |
+----+------+
| 9  | 2017 |
+----+------+
| 8  | 2017 |
+----+------+
| 7  | 2017 |
+----+------+
| 6  | 2017 |
+----+------+
| 5  | 2017 |
+----+------+
| 4  | 2017 |
+----+------+
| 3  | 2017 |
+----+------+
| 2  | 2017 |
+----+------+
| 1  | 2017 |
+----+------+
| 12 | 2016 |
+----+------+
| 11 | 2016 |
+----+------+
| 10 | 2016 |
+----+------+
| 9  | 2016 |
+----+------+
| 8  | 2016 |
+----+------+
| 7  | 2016 |
+----+------+
| 6  | 2016 |
+----+------+
| 5  | 2016 |
+----+------+
| 4  | 2016 |
+----+------+
| 3  | 2016 |
+----+------+
| 2  | 2016 |
+----+------+
| 1  | 2016 |
+----+------+
| 12 | 2015 |
+----+------+
| 11 | 2015 |
+----+------+
| 10 | 2015 |
+----+------+
| 9  | 2015 |
+----+------+
| 8  | 2015 |
+----+------+
| 7  | 2015 |
+----+------+
| 6  | 2015 |
+----+------+
| 5  | 2015 |
+----+------+
| 4  | 2015 |
+----+------+
| 3  | 2015 |
+----+------+
| 2  | 2015 |
+----+------+
| 1  | 2015 |
+----+------+
| 12 | 2014 |
+----+------+
| 11 | 2014 |
+----+------+
| 10 | 2014 |
+----+------+
| 9  | 2014 |
+----+------+
| 8  | 2014 |
+----+------+
| 7  | 2014 |
+----+------+
| 6  | 2014 |
+----+------+
| 5  | 2014 |
+----+------+
| 4  | 2014 |
+----+------+
| 3  | 2014 |
+----+------+
| 2  | 2014 |
+----+------+
| 1  | 2014 |
+----+------+
| 12 | 2013 |
+----+------+
| 11 | 2013 |
+----+------+
| 10 | 2013 |
+----+------+
| 9  | 2013 |
+----+------+
| 8  | 2013 |
+----+------+
| 7  | 2013 |
+----+------+
| 6  | 2013 |
+----+------+
| 5  | 2013 |
+----+------+
| 4  | 2013 |
+----+------+
| 3  | 2013 |
+----+------+
| 2  | 2013 |
+----+------+
| 12 | 2012 |
+----+------+
| 12 | 2011 |
+----+------+
| 12 | 2010 |
+----+------+
| 12 | 2009 |
+----+------+
| 12 | 2008 |
+----+------+
| 12 | 2007 |
+----+------+
| 12 | 2006 |
+----+------+
| 12 | 2005 |
+----+------+

我正在考虑用几种方法来做到这一点，也许是交叉连接，或者是一些 while 循环——但它们看起来效率很低。我不认为它是最难做的，但我想为它编写 'good' 代码。我已经思考 "best" 方式好一段时间了。

不幸的是，我对数据如何到达我没有任何发言权，这是我需要提供的 return，否则我将使用日期数据类型和日期函数。

Answer 1

这使用递归 CTE。如果您的环境允许，我自然会使用 dimension table。这是一种方式，但最好的方式当然是有争议的。

declare @year int = 2018
declare @month int = 1



;with cte as(
    select [Month], [Year] = 2005
    from  (values(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12)) as X([Month]) 
    union all
    select [Month], [Year] + 1
    from cte
    where [Year] < @year
),

someYears as(
select [Years] = 2005
union all
select [Years] + 1
from someYears
where [Years] < @year),

t as(
select  *, RN = row_number() over (order by [Year] desc, [Month] desc) 
from cte
where [Month] <= case when [Year] = @year then @month else 13 end)


select [Month], [Year]
from t
where RN <= 60
union 
select 12, [Years]
from (select [Years] from someYears) x
where x.Years < @year
order by [Year] desc, [Month] desc

Answer 2

我的偏好是在 table 中坚持这一点。基本上，创建一个尽可能远的 table，然后让它进入大约 50 年后的未来。然后，您的函数可以 select 来自 table.

table:

create table MonthDim
(
PK_ID int identity(1, 1) primary key,
Month int,
Year int,
Dt date --set this to the first of each month
)

查询：

select *
from MonthDim
where dt >= dateadd(month, -60, datefromparts(@year, @month, 1))
    and dt <= datefromparts(@year, @month, 1)

union all

select *
from MonthDim
where dt < dateadd(month, -60, datefromparts(@year, @month, 1))
    and Month = 12

这样做的额外好处是，由于它是持久化的，如果您需要保存下游生成的任何数据，您可以在此处引用主键。它也更容易理解正在发生的事情。

我如何return 60组月份和年份数据在一个月份和年份列中？

How do I return 60 sets of month and year data in a month and year column?

sql-server

sql-server-2017