假设我有一个包含以下列的表 (开始,结束,间隔)
有没有什么方法可以获得Start,End之间的所有值以及每一行的间隔? 请注意,(Start,End,Interval)表中的行不止一行,但它们不应重叠。
如果可能,没有循环/游标/临时表/变量表。
示例数据
Start End Interval 1 3 1 9 12 1 16 20 2
期望的结果:
Result 1 2 3 9 10 11 12 16 18 20
答案 0 :(得分:4)
这是递归公用表表达式的一个很好的用例:
;with cte as (
select [Start] as Result, [End], [Interval]
from Table1
union all
select Result + [Interval], [End], [Interval]
from cte
where Result + [Interval] <= [End]
)
select Result
from cte
order by Result
<强> sql fiddle demo 强>
答案 1 :(得分:2)
你可以这样做
WITH tally AS (
SELECT 0 n
UNION ALL
SELECT n + 1 FROM tally WHERE n < 100 -- adjust 100 to a max possible value for (end - start) / interval
)
SELECT start + n * [interval] result
FROM Table1 t CROSS JOIN tally n
WHERE n.n <= (t.[end] - t.start) / t.[interval]
ORDER BY result
注意:如果您执行了大量此类查询,则可以考虑将带有主键的持久数字表tally
替换为递归CTE tally
n
1}}列。
输出:
| RESULT | |--------| | 1 | | 2 | | 3 | | 9 | | 10 | | 11 | | 12 | | 16 | | 18 | | 20 |
这是 SQLFiddle 演示
答案 2 :(得分:1)
我知道你接受了答案,我认为这也是正确的。
<强> Fiddle demo 1 强>
select x.number
from master..spt_values x cross join table1 t
where x.type='p' and x.number between t.[start] and t.[end]
and x.number % t.[interval] = 0
结果:
| NUMBER |
|--------|
| 1 |
| 2 |
| 3 |
| 9 |
| 10 |
| 11 |
| 12 |
| 16 |
| 18 |
| 20 |
编辑:如果您想要无限数量,请尝试此方法并根据需要交叉加入更多数字表。这个例子最高可达9999。
<强> Fiddle demo 2 强>:
;WITH Digits AS (
select Digit
from ( values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9))
AS t(Digit))
,Numbers AS (
select u.Digit + t.Digit*10 + h.Digit*100 + th.Digit*1000 as number
from Digits u
cross join Digits t
cross join Digits h
cross join Digits th
--Add more cross joins as required
)
Select number
From Numbers x cross join table1 t
where x.number between t.[start] and t.[end]
and x.number % t.[interval] = 0;
Order by number