查找跨越多列的數字範圍的差距

Question

我試圖找出數字範圍（SQL Server）中的差距。我的情況是低於...

ID Start End 
1  1  4 
2  1  6 
3  2  4 
4  8  10 
5  13  14 

Visual 
------------------------------- 
1-2-3-4 
1-2-3-4-5-6 
    2-3-4 
      - -8-9-10 
        - - -13-14 

這樣做的結果可能是沿着線的東西：

Table 
------------------------------- 
ID Start End Gap 
4  8  10 -1 
5  13  14 -2 

最後，我想有間隙的範圍，但我應該能夠人物說出來上面......

Missing 
7 
11-12 

我想出了是不是太緩慢或不佔範圍的重疊（例如ID 2）

解決方案

CREATE TABLE #Docs (
    [Rank] INT, --DENSE_RANK() OVER(ORDER BY BegProd) 
    ControlNumber BIGINT, 
    BegProd INT, 
    EndProd INT 
) 

SELECT 
    T1.ControlNumber, 
    T1.BegProd, 
    T1.EndProd, 
    MAX(T2.EndProd) AS [PreviousEndProd], 
    [Gap] = T1.BegProd - MAX(T2.EndProd) - 1 
FROM #Docs T1 
INNER JOIN #Docs T2 
    ON T1.[Rank] = T2.[Rank] + 1 
    AND T1.EndProd > T2.EndProd 
GROUP BY T1.ControlNumber, T1.BegProd, T1.EndProd 
HAVING T1.BegProd - MAX(T2.EndProd) > 1 

有200多萬行此表和1比1十億

編輯 固定「失蹤」表的範圍跨度。 間隔列表示在起始數字之前有多少間隙。 （如缺少＃7是1號）

Answer 1

試試這個：

create table #docs(id int, start int, [end] int) 
insert #docs values(1,1,4),(2,1,6),(3,2,4),(4,8,10),(5,13,14) 

;with a as 
(
select start, dense_rank() over (order by start) rn 
from #docs t where not exists (select 1 from #docs where t.start > start and t.start < [end]) 
group by start 
), b as 
(
select [end], dense_rank() over (order by [end]) rn 
from #docs t where not exists (select 1 from #docs where t.[end] > start and t.[end] < [end]) 
group by [end] 
) 
select 
case when a.[start]= b.[end]+2 then cast(a.start-1 as varchar(21)) 
else cast(b.[end]+1 as varchar(10)) +'-' + cast(a.start - 1 as varchar(10)) end missing 
from a join b on a.rn - 1 = b.rn 
and a.[start] <> b.[end] + 1 

結果：

Missing 
7 
11-12