[英]use recursive common table expressions to find consecutive no.s from two tables
我有以下表格:
Actual Optional
------ --------
4 3
13 6
20 7
26 14
19
21
27
28
我要做的是选择:
1)“实际”表中的所有值。
2)如果它们形成具有“实际”表值的连续系列,则从“可选”表中选择值
预期的结果是:
Answer
------
4
13
20
26
3 --because it is consecutive to 4 (i.e 3=4-1)
14 --14=13+1
19 --19=20-1
21 --21=20+1
27 --27=26+1
28 --this is the important case.28 is not consecutive to 26 but 27
--is consecutive to 26 and 26,27,28 together form a series.
我使用递归cte编写了一个查询但是它永远循环并且在递归达到100级后失败。 我面临的问题是27场比赛26场比赛,27场比赛27场比赛27场比赛27场比赛28场比赛27场比赛......(永远)
这是我写的查询:
with recurcte as
(
select num as one,num as two from actual
union all
select opt.num as one,cte.two as two
from recurcte cte join optional opt
on opt.num+1=cte.one or opt.num-1=cte.one
)select * from recurcte
;WITH Combined
AS (SELECT 1 AS Actual, N
FROM (VALUES(4),
(13),
(20),
(26)) Actual(N)
UNION ALL
SELECT 0 AS Actual, N
FROM (VALUES(3),
(6),
(7),
(14),
(19),
(21),
(27),
(28)) Optional (N)),
T1
AS (SELECT *,
N - DENSE_RANK() OVER (ORDER BY N) AS Grp
FROM Combined),
T2
AS (SELECT *,
MAX(Actual) OVER (PARTITION BY Grp) AS HasActual
FROM T1)
SELECT DISTINCT N
FROM T2
WHERE HasActual = 1
此CTE将为您提供所需的数据。 这不需要递归。
declare @Actual table (i int)
declare @Optional table (i int)
insert into @Actual
select 4 union select 13 union select 20 union select 26
insert into @Optional
select 3 union select 6 union select 7 union select 14 union select 19
union select 21 union select 27 union select 28
;with rownum as (
select *, ROW_NUMBER() OVER (ORDER BY i) as 'RN'
from (
select
i, 'A' as 'Source'
from
@Actual
union
select
i, 'O'
from
@Optional
) a
)
select distinct
d.i
from
rownum a
inner join rownum d
on a.i - d.i = a.rn - d.rn
where
a.source = 'A'
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