我正在寻找一种在单个select语句中组合连续数字范围的简洁方法.
说我的表有这些记录:
first_number last_number
0 9
10 19
20 29
40 49
50 59
70 79
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然后输出如下:
first_number last_number
0 29
40 59
70 79
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这就是我想出来的:
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
;
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这很好用.只是,对于如此小的任务来说,它看起来如此复杂.我很想知道还有比我更直接的方式.
请尝试:
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
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