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