选取连续的记录
表格如下：
id se
24022
24021
16093
16012
16011
15051
09041
06016
05083
.. ..
.. ..
se 是不确定的tinyint,不一定是从1开始。。想选取有连续id/se 的记录（> =2条），
想得到下面的结果集合：
2402 2
2402 1
1601 2
1602 1
[解决办法]
create table tbl
(
id varchar(4),
se int
)
insert into tbl
select '2402 ',2
union all select '2402 ',1
union all select '1609 ',3
union all select '1601 ',2
union all select '1601 ',1
union all select '1505 ',1
union all select '0904 ',1
select * from tbl a
where exists (select 1 from tbl where a.id =id group by id having count(1)> 1 )
[解决办法]
declare @t table(id varchar(10),se int)
insert @t select '2402 ',2
union all select '2402 ',1
union all select '1609 ',3
union all select '1601 ',2
union all select '1601 ',1
union all select '1505 ',1
union all select '0904 ',1
union all select '0601 ',6
union all select '0508 ',3
select * from @t a where exists(select 1 from @t where id=a.id and (se=a.se-1 or se=a.se+1))
--结果
id se
---------- -----------
2402 2
2402 1
1601 2
1601 1
（所影响的行数为 4 行）
[解决办法]
declare @t table(id varchar(10),se int)
insert @t select '2402 ',2
union all select '2402 ',1
union all select '1609 ',3
union all select '1601 ',2
union all select '1601 ',1
union all select '1505 ',1
union all select '0904 ',1
union all select '0601 ',6
union all select '0508 ',3
select *
from @t a
where exists(
select 1
from
@t
where id=a.id and abs(se - a.se) = 1
)
order by a.id,a.se
/*
id se
---------- -----------
1601 1
1601 2
2402 1
2402 2
*/
