如何检查某个范围是否被一组范围 完全覆盖 。在以下示例中:
WITH ranges(id, a, b) AS (
SELECT 1, 0, 40 UNION
SELECT 2, 40, 60 UNION
SELECT 3, 80, 100 UNION
SELECT 4, 10, 30
), tests(id, a, b) AS (
SELECT 1, 10, 90 UNION
SELECT 2, 10, 60
)
SELECT *
FROM tests
WHERE -- ?
10, 60
,因为所有内容都被0, 40
和40, 60
(以及10, 30
)覆盖10, 90
,因为它暴露在60, 80
之间假定a
是包含性的,而b
是排除性的,即值40
属于[40, 60)
而不是[0, 40)
。范围可以包含间隙和各种重叠。
实际问题涉及日期和时间数据,但日期只是数字。我正在使用SQL Server,但首选通用解决方案。
答案 0 :(得分:2)
这是类似于Thorsten的递归解决方案。只是提供另一个示例。
WITH ranges(id, a, b) AS (
SELECT 1, 0, 40 UNION
SELECT 2, 40, 60 UNION
SELECT 3, 80, 100 UNION
SELECT 4, 10, 30
), tests(id, a, b) AS
(
SELECT 1 as id, 10 as a, 90 as b
UNION
SELECT 2, 10, 60
), rangeFinder(a, b, ra, rfb) AS
(
SELECT a, b, 0 AS ra, 0 AS rfb
FROM ranges AS r
UNION ALL
SELECT rangeFinder.a, ranges.b, ranges.a, rangeFinder.b
FROM ranges
JOIN rangeFinder
ON ranges.b > rangeFinder.b
AND ranges.a <=rangeFinder.b
), islands(a, b) AS
(
SELECT a, b
FROM rangeFinder
WHERE a NOT IN (SELECT ra FROM rangeFinder)
AND b NOT IN (SELECT rfb FROM rangeFinder)
)
SELECT t.id, t.a, t.b FROM
tests t
JOIN islands i
ON t.a >= i.a
AND t.b <= i.b
答案 1 :(得分:1)
您想要一个递归查询来查找实际范围(在您的情况下为0到60和80到100)。我们将从给出的范围开始,然后寻找扩展这些范围的范围。最后,我们坚持使用最宽的范围(例如,可以将范围10到30扩展到0到40,然后再扩展到0到60,因此我们将最宽的范围保持在0到60)。
with wider_ranges(a, b, grp) as
(
select a, b, id from ranges
union all
select
case when r1.a < r2.a then r1.a else r2.a end,
case when r1.b > r2.b then r1.b else r2.b end,
r1.grp
from wider_ranges r1
join ranges r2 on (r2.a < r1.a and r2.b >= r1.a)
or (r2.b > r1.b and r2.a <= r1.b)
)
, real_ranges(a, b) as
(
select distinct min(a), max(b)
from wider_ranges
group by grp
)
select *
from tests
where exists
(
select *
from real_ranges
where tests.a >= real_ranges.a and tests.b <= real_ranges.b
);
上个月的演示:http://rextester.com/BDJA16583
根据要求,它可以在SQL Server中工作,但是它是标准SQL,因此它应该可以在所有具有递归查询的DBMS中使用。
答案 2 :(得分:0)
这是解决方案的一般形式。该想法是要执行以下操作:
这是基于以下操作:不在范围内的点将成为任一表中包含的数字之一:
with tc as (
select t.test, r.candidate
from tests t join
(select r.a as candidate from ranges union all
select r.b from ranges
) r
on r.candiate >= t.a and r.candidate < t.b
union all
select t.test, t.a
from tests t
union all
select t.test, t.b
from tests t
)
select distinct tc.test
from tc
where not exists (select 1
from ranges r
where tc.candidate >= r.a and tc.candidate < r.b
);
由于范围包括第一项,因此您实际上不必进行检查。因此可以减少候选人的名单:
with tc as (
select t.test, r.candidate
from tests t join
(select r.b as candidate from ranges
) r
on r.candidate >= t.a and r.r < t.b
union all
select t.test, t.a
from tests t
union all
select t.test, t.b
from tests t
)
答案 3 :(得分:0)
如接受的答案中所述,解决方案是将重叠范围合并在一起,然后确定合并范围之一内是否存在测试范围。
除了连接和/或递归,您还可以将sorting approach与窗口函数一起使用来合并重叠范围:
WITH ranges(id, a, b) AS (
SELECT 1, 0, 40 UNION
SELECT 2, 40, 60 UNION
SELECT 3, 80, 100 UNION
SELECT 4, 10, 30
), tests(id, a, b) AS (
SELECT 1, 10, 90 UNION
SELECT 2, 10, 60
), ranges_chg AS (
SELECT *, CASE WHEN MAX(b) OVER (ORDER BY a ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING) >= a THEN 0 ELSE 1 END AS chg
FROM ranges
), ranges_grp AS(
SELECT *, SUM(chg) OVER (ORDER BY a) AS grp
FROM ranges_chg
), merged_ranges AS (
SELECT MIN(a) AS a, MAX(b) AS b
FROM ranges_grp
GROUP BY grp
)
SELECT *
FROM tests
WHERE EXISTS (
SELECT 1
FROM merged_ranges
WHERE merged_ranges.a <= tests.a AND tests.b <= merged_ranges.b
)
结果和Fiddle。
| id | a | b |
|----|----|----|
| 2 | 10 | 60 |
range_grp
CTE中的数据将使您了解其工作方式:
| id | a | b | max b over... | chg | grp |
|----|----|-----|---------------|-----|-----|
| 1 | 0 | 40 | NULL | 1 | 1 |
| 4 | 10 | 30 | 40 | 0 | 1 |
| 2 | 40 | 60 | 40 | 0 | 1 |
| 3 | 80 | 100 | 60 | 1 | 2 |