找到連續的數字

Question

好的 - 所以我搜索了這個網絡，沒有一個我發現的例子和我的一模一樣。

我有一個5列和數千行的表。 我需要在每行中找到連續的數字。我需要使用以下

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) 

這就是我要的列移動到行3次所示的查詢的情況下結束了，但我不知道現在該怎麼檢查。

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 

謝謝你的幫助！

@TimBiegeleise，更新： 所以我發現這對谷歌的差距&羣島

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; 

這是我更新查詢轉換的列行（但需要2條語句，我更願意有1 ）和實施島嶼部分 - 但我不明白如何給我的結果我需要（見上文）。下面我顯示原始行數據和結果。

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 

結果我用上面的查詢得到：

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

幫助:-)！

Answer 1

好的，我明白了。對於上述行，這個查詢返回值爲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