SQL - 如何根据现有数据集中的日期范围为每个月生成行？

Question

假设我有一个数据集:

rowID |  dateStart  |   dateEnd  | Year | Month
121   | 2013-10-03  | 2013-12-03 | NULL | NULL
143   | 2013-12-11  | 2014-03-11 | NULL | NULL
322   | 2014-01-02  | 2014-02-11 | NULL | NULL

我希望sql根据dateStart和dateEnd生成以下数据源.请注意年份和月份分组.

rowID |  dateStart  |   dateEnd  | Year | Month
121   | 2013-10-03  | 2013-12-03 | 2013 |   10
121   | 2013-10-03  | 2013-12-03 | 2013 |   11
121   | 2013-10-03  | 2013-12-03 | 2013 |   12
143   | 2013-12-11  | 2014-03-11 | 2013 |   12
143   | 2013-12-11  | 2014-03-11 | 2014 |    1
143   | 2013-12-11  | 2014-03-11 | 2014 |    2
143   | 2013-12-11  | 2014-03-11 | 2014 |    3
322   | 2014-01-02  | 2014-02-11 | 2014 |    1
322   | 2014-01-02  | 2014-02-11 | 2014 |    2

我很难缠绕这个.有任何想法吗？

Answer 1

通过创建整数列表然后使用它来递增日期,我发现最容易解决这些问题.这是一个例子:

with nums as (
      select 0 as n
      union all
      select n + 1 as n
      from nums
      where n < 11
     )
select rowid, datestart, dateend,
       year(dateadd(month, n.n, datestart)) as yr,
       month(dateadd(month, n.n, datestart)) as mon
from table t join
     nums n
     on dateadd(month, n.n - 1, datestart) <= dateend;