使用Oracle SQL和JOINS连续编号的间隔

Question

我试图以每种形式显示冰岛连续海拔的间隔：

ELEVATION   
         0        27
         29       33
         35
         37       40
         42       46
         48
         51       63

目前，我只是设法跟踪差距，我有一些想法：

  

如果（columnB中的值的计数直到columnC =＆＃39; GAP＆＃39;）等于   columnB的值，其中columnC =＆＃39; GAP＆＃39;然后我们有一个连续的间隔   在columnA和column B的值之间

有人能给我一些提示吗？

当前代码

with x as (
    SELECT distinct elevation
    FROM CITIES
    WHERE iso = 'IS' AND iso IS NOT NULL
),y as (
    SELECT a.ELEVATION as "A",B.ELEVATION as "B",C.ELEVATION as "C"
FROM x a
   JOIN x b ON b.ELEVATION > a.ELEVATION
  LEFT JOIN x c ON c.ELEVATION > b.ELEVATION AND c.ELEVATION < b.ELEVATION + 2
)select y.A,y.B,y.C,case when y.C is null then 'GAP' else ' ' end  GAPZ from y
order by 1,2

输出

 A  B   C   GAP
------------------
0   1   2    
0   2   3    
0   3   4    
...
...
...  
0   25  26   
0   26  27   
0   27      GAP
0   29  30   
0   30  31   
0   31  32   
0   32  33   
0   33      GAP
0   35      GAP
0   37  38   
0   38  39   
0   39  40   
0   40      GAP
0   42  43   
0   43  44   
0   44  45   
0   45  46   
0   46      GAP
0   48      GAP
0   51  52   
0   52  53   
...
...
...
0   61  62   
0   62  63   
0   63      GAP
0   65  66   
0   66  67   
0   67  68   
0   68  69   
0   69      GAP
0   71  72   
...
...
...

Answer 1

你已经到了一半！您需要标记间隙的任一端，然后仅选择那些行。

Select x, 
            Case lead(x,x) over(order by x)
                 When x+1 then null
                 Else x
             End as endpoint,
             Case lag(x,x) over(order by x)
                  When x-1 then null
                  Else x
              End as startpoint
  From table

这将显示行是开始还是端点。我们称之为Q1。现在我们只需从中选择我们需要的东西。

Select Q1.startpoint,
            (Select min(endpoint)
               From Q1 as endp
             Where endp.endpoint >= Q1.x) as endpoint
 From Q1
 Where Q1.startpoint is not null