（Sqlserver）sql求连续问题

题目一：
create table etltable(
name varchar(20) ,
seq int,
money int);
 
create table etltarget (
name varchar(20),
min_s int,
max_s int,
sum_money int);
 
insert into etltable values
('A',1,100),
('A',2,200),
('A',3,300),
('A',8,400),
('A',9,500),
('B',1,100),
('B',2,500),
('B',5,600);
 
目标表结果应该是这样
A 1 3 600
A 8 9 900
B 1 2 600
B 5 5 600
 
解答：
	select name,
	min(seq) as min_s,
	max(seq) as max_s,
	sum(money) as money
	from (
	select
	*,
	seq-row_number() over (partition by name order by name) as rn2--连续的差值会相同
	from etltable
	) a group by name,rn2--连续的差值相同，按照这个分组即可
	order by name

　　

题目二：
with aa
as (
select 1 uid, '2015-6-1 8:20:00' as login_time union all
select 1 uid, '2015-6-2 7:20:05' as login_time union all
select 1 uid, '2015-6-3 21:20:30' as login_time union all
select 2 uid, '2015-6-1 8:10:00' as login_time union all
select 2 uid, '2015-6-3 8:20:00' as login_time union all
select 2 uid, '2015-6-4 18:20:00' as login_time union all
select 1 uid, '2015-6-5 9:20:00' as login_time union all
select 1 uid, '2015-6-6 16:20:00' as login_time union all
select 3 uid, '2015-6-1 8:20:00' as login_time union all
select 1 uid, '2015-6-7 12:20:00' as login_time union all
select 1 uid, '2015-6-8 23:20:00' as login_time union all
select 3 uid, '2015-6-2 8:20:00' as login_time union all
select 3 uid, '2015-6-3 2:20:00' as login_time
 
)
 
select
uid,min(ds) as min_s,max(ds) as max_s,sum(1) cnt
from (
select * ,
day(login_time) ds,
day(login_time)-row_number() over (partition by uid order by login_time) as rn
from aa
) a
group by uid,rn