4

我正在寻找一种巧妙的方法来在单个选择语句中组合连续的数字范围。

假设我的表有这些记录：

first_number    last_number
0   9
10  19
20  29
40  49
50  59
70  79

那么输出应该如下：

first_number    last_number
0   29
40  59
70  79

这就是我想出的：

select first_number, last_number_of_range
from
(
  select 
    first_number, is_continuing, is_continued,
    nvl(lead (last_number,1,null) over (order by first_number), last_number) as last_number_of_range
  from
  (
    select *
    from
    (
      select first_number, last_number, 
       case when lag (last_number,1,null) over (order by first_number) + 1 = first_number then 1 else 0 end as is_continuing, 
       case when lead (first_number,1,null) over (order by last_number) - 1 = last_number then 1 else 0 end as is_continued
      from 
      (
        select 0 as first_number, 9 as last_number from dual
        union all
        select 10 as first_number, 19 as last_number from dual
        union all
        select 20 as first_number, 29 as last_number from dual
        union all
        select 40 as first_number, 49 as last_number from dual
        union all
        select 50 as first_number, 59 as last_number from dual
        union all
        select 70 as first_number, 79 as last_number from dual
      )
    )
    where is_continuing = 0 or is_continued = 0 -- remove all but first and last of consecutive records
  )
)
where is_continuing = 0 -- now at last remove those records that gave us the last_number_of_range
;

这工作正常。只是，对于这么小的任务，它看起来太复杂了。我很想知道是否有比我的更直接的方法。

4

3 回答 3

3

这是另一种方法，它将为您提供所需的输出。

select min(first_number)  as first_number
     , max(last_number)   as last_number
 from ( 
        select first_number
              , last_number
              , sum(grp) over(order by first_number) as grp
           from ( select first_number
                       , last_number
                       , case
                           when first_number <>
                                   lag(last_number) 
                                     over(order by first_number) + 1
                           then 1
                           else 0
                         end as grp
                     from t1 )
       )
group by grp
order by 1

结果：

FIRST_NUMBER LAST_NUMBER
------------ -----------
           0          29
          40          59
          70          79

SQLFiddle 演示

于 2013-09-20T10:30:46.383 回答
2

请试试：

with T1 as (
  select 
    row_number() over (order by first_number) RNum, 
    first_number, 
    last_number 
  From yourtable
)
,T (RNUM, first_number, last_number, CNT) as (
    select T1.*, 1 CNT from T1 where RNum=1
    union all
    SELECT b.RNUM, b.first_number, b.last_number, (case when b.first_number=T.last_number+1 then t.CNT 
                        else T.CNT+1 end) CNT
    from T1 b INNER JOIN T on b.RNum=T.RNum+1
)
select 
   min(first_number) as first_number,
    max(last_number) as last_number
From T group by T.CNT

SQL小提琴演示

于 2013-09-20T10:31:14.977 回答
0

一个旧线程，但我有另一个解决方案......在我的测试中运行良好，而且看起来很简单。

未注释的一个：

SELECT 
  MAX(r2.first_number) AS first_number, 
  r1.last_number 
FROM <table> AS r1
INNER JOIN <table> AS r2 ON 1=1
  AND r2.first_number < r1.last_number
LEFT JOIN <table> AS r3 ON 1=1
  AND r3.last_number  >= r1.last_number + 1
  AND r3.first_number <= r1.last_number + 1
LEFT JOIN <table> AS r4 ON 1=1
  AND r4.last_number  >= r2.first_number - 1
  AND r4.first_number <= r2.first_number - 1
WHERE 1=1
  and r3.first_number IS NULL
  and r4.first_number IS NULL
GROUP BY r1.last_number
;

评论一：

SELECT 
  -- getting a first_number 
  -- lesser than the 
  -- maximum first_number available
  -- makes this maximum one
  -- an overlap...
  MAX(r2.first_number) first_number, 
  r1.last_number 
FROM <table> r1
INNER JOIN <table> r2 ON 1=1
  -- join the data so 
  -- for  a last_number  in r1 you can
  -- find a first_number in r2 that is
  -- lesser than its value, as 
  -- no first_number should be
  -- greater than (or equals?) a last_number
  AND r2.first_number < r1.last_number
LEFT JOIN <table> r3 ON 1=1
  -- matches when another range 
  -- overlaps last_number from r1
  AND r3.last_number  >= r1.last_number + 1
  AND r3.first_number <= r1.last_number + 1
LEFT JOIN <table> r4 ON 1=1
  -- matches when another range 
  -- overlaps first_number from r2
  AND r4.last_number  >= r2.first_number - 1
  AND r4.first_number <= r2.first_number - 1
WHERE 1=1
  -- get rows 
  -- with no overlaps on last_number and
  -- with no overlaps on first_number
  and r3.first_number IS NULL
  and r4.first_number IS NULL
GROUP BY r1.last_number
;
于 2016-09-25T18:55:07.307 回答