Question

好的 - 所以我在互联网上搜索了这个，我找到的所有例子都不像我的。

我有一个包含5列和数千行的表。我需要在每一行中找到连续的数字。我需要针对下面显示的情况进行3次查询

n1   n2   n3   n4   n5
=======================
 1     3    4    6    9   = should result in 1 (when checking for pairs)
 1     3    4    5    9   = should result in 1 (when checking for triplets)
 1     2    5    8    9   = should result in 1 (when checking for double pairs)

这就是我必须将列移动到行中，但我不知道如何检查这些。

select n1 from (
select n1 from myTable where Id  = 1
union all select n2 from myTable where Id = 1
union all select n3 from myTable where Id = 1
union all select n4 from myTable where Id = 1
union all select n5 from myTable where Id = 1
) t
order by n1

感谢您的帮助！

@TimBiegeleise，更新：所以我在Google上找到了Gaps＆amp;岛屿：

SELECT ID, StartSeqNo=MIN(SeqNo), EndSeqNo=MAX(SeqNo)
FROM (
SELECT ID, SeqNo
    ,rn=SeqNo-ROW_NUMBER() OVER (PARTITION BY ID ORDER BY SeqNo)
FROM dbo.GapsIslands) a
GROUP BY ID, rn;

这是我更新的查询将列转换为行（但它需要2个语句，我更确切地说有1个）并实现岛部分 - 但我不明白如何给我结果我需要的（见上文）。下面我显示原始行数据和结果。

select n1, IDENTITY (INT, 1, 1) AS ID 
into #test
from (
select n1 from myTable where Id  = 8
union all select n2 from myTable where Id = 8
union all select n3 from myTable where Id = 8
union all select n4 from myTable where Id = 8
union all select n5 from myTable where Id = 8
) as t
order by n1

SELECT ID, StartSeqNo=MIN(n1), EndSeqNo=MAX(n1)
FROM (
SELECT ID, n1
    ,rn=n1-ROW_NUMBER() OVER (PARTITION BY ID ORDER BY n1)
FROM #test) a
GROUP BY ID, rn

drop table #test

original row - should return 1 (when checking for "pair"/consecutive numbers
n1   n2   n3   n4   n5
=======================
31   27   28   36   12

我得到上述查询的结果：

    StartSeqNo  EndSeqNo
1   12          12
2   27          27
3   28          28
4   31          31
5   36          36

帮助:-)！

Answer 1

好的，我明白了。此查询为上述行

返回值1

select COUNT(*) as pairs 
from (
SELECT StartSeqNo=MIN(n1), EndSeqNo=MAX(n1)
    FROM (
        SELECT n1, rn=n1-ROW_NUMBER() OVER (ORDER BY n1)
            from (
                select n1 from myTable where Id  = 8
                union all select n2 from myTable where Id = 8
                union all select n3 from myTable where Id = 8
                union all select n4 from myTable where Id = 8
                union all select n5 from myTable where Id = 8
            ) t
    ) x
GROUP BY rn
) z
where StartSeqNo+1 = EndSeqNo

找到连续的数字

1 个答案: