SQL查找大小为n的连续区间

数据准备

 1 create table sequence
 2 (
 3 seq int not null primary key
 4 );
 5 
 6 insert into values(3);
 7 insert into values(4);
 8 insert into values(5);
 9 insert into values(6);
10 insert into values(7);
11 insert into values(8);
12 insert into values(10);
13 insert into values(11);
14 insert into values(15);
15 insert into values(16);
16 insert into values(17);
17 insert into values(20);
18 insert into values(22);
19 insert into values(23);
20 insert into values(24);
21 insert into values(28);
22 insert into values(30);
23 insert into values(31);

如何要查找长度为4的区间首数和尾数，即@n=4

解决方案一：

select S1.seq as start,S1.seq+@n-1 as end
from sequence as S1 cross join sequence as S2 cross join sequence as S3
where S2.seq = S1.seq+@n-1
group by S1.seq
having sum(case when S3.seq >S1.seq and S3.seq<=S2.seq then 1 else 0 end)=3;

View Code

此方案用到了自联接，where条件决定了起点和终点，group by 子句用于生成递归集合，还需要一个能够描述区间内部各个点需要满足的条件，为此增加一个自联接表S3，having子句针对子集内元素个数作出约束，符合约束的集合将被取出。

1 +-------+-----+
2 | start | end |
3 +-------+-----+
4 |     3  |   6 |
5 |     4  |   7 |
6 |     5  |   8 |
7 |    15 |  18 |
8 +-------+-----+

结果