Question

仅为数字+1的范围选择范围（最小值，最大值）

有一张桌子：

col_number
1
2
3
4
9
10
12

必须如此获取数据：

Answer 1

这称为间隙和孤岛问题。您可以通过从值中减去行号来获得组键来解决此问题：

select min(col), max(col)
from
(
  select
    col,
    col - row_number() over (order by col) as grp
  from mytable
) 
group by grp
order by min(col);

Answer 2

在Oracle 12.1或更高版本中，您可以像这样使用match_recognize：

with
  my_table(col_number) as (
    select  1 from dual union all
    select  2 from dual union all
    select  3 from dual union all
    select  4 from dual union all
    select  9 from dual union all
    select 10 from dual union all
    select 12 from dual
  )
-- end of sample data (for testing only, not part of the solution)
select min_number, max_number
from   my_table
match_recognize(
  order by col_number
  measures first(col_number) as min_number, last(col_number) as max_number
  pattern  (a b*)
  define b as col_number = prev(col_number) + 1
);

MIN_NUMBER MAX_NUMBER
---------- ----------
         1          4
         9         10
        12         12

Answer 3

在Oracle中，您实际上可以在没有窗口功能的情况下执行此操作：

-DtestApp=app2

Answer 4

没有窗口功能：

WITH 
nmin AS (
  SELECT col_number FROM numbers t WHERE NOT EXISTS 
    (SELECT col_number FROM numbers WHERE numbers.col_number = t.col_number - 1)),
nmax AS (
  SELECT col_number FROM numbers t WHERE NOT EXISTS 
    (SELECT col_number FROM numbers WHERE numbers.col_number = t.col_number + 1))
SELECT 
  nmin.col_number minnumber, 
  (SELECT MIN(nmax.col_number) FROM nmax WHERE nmax.col_number >= nmin.col_number) maxnumber
FROM nmin
ORDER BY nmin.col_number;

仅为数字+1选择范围（最小，最大）

4 个答案: