---找出促銷(xiāo)活動(dòng)中銷(xiāo)售額最高的職員
---你剛在一家服裝銷(xiāo)售公司中找到了一份工作，此時(shí)經(jīng)理要求你根據(jù)數(shù)據(jù)庫(kù)中的兩張表得到促銷(xiāo)活動(dòng)銷(xiāo)售額最高的銷(xiāo)售員
---1.一張是促銷(xiāo)活動(dòng)表
---2.一張是銷(xiāo)售客列表
create table Promotions
(
activity nvarchar(30),
sdate datetime,
edate datetime
)
insert Promotions
select '五一促銷(xiāo)活動(dòng)','2011-5-1','2011-5-7'
union
select '十一促銷(xiāo)活動(dòng)','2011-10-1','2011-10-7'
union
select 'OA專(zhuān)場(chǎng)活動(dòng)','2011-6-1','2011-6-7'
go
create table sales
(
id int not null,
name nvarchar(20),
saledate datetime,
price money
)
go
insert sales
select 1,'王五','2011-5-1',1000 union
select 1,'王五','2011-5-2',2000 union
select 1,'王五','2011-5-3',3000 union
select 1,'王五','2011-5-4',4000 union
select 1,'張三','2011-5-1',1000 union
select 1,'張三','2011-5-3',2000 union
select 1,'張三','2011-5-4',4000 union
select 1,'李四','2011-5-6',1000 union
select 1,'趙六','2011-5-5',1000 union
select 1,'錢(qián)七','2011-5-8',1000 union

select 1,'孫五','2011-6-1',1000 union
select 1,'孫五','2011-6-2',2000 union
select 1,'王五','2011-6-3',3000 union
select 1,'孫五','2011-6-4',4000 union
select 1,'張三','2011-6-1',11000 union
select 1,'張三','2011-6-3',20000 union
select 1,'張三','2011-6-4',4000 union
select 1,'李四','2011-6-6',1000 union
select 1,'趙六','2011-6-5',1000 union
select 1,'錢(qián)七','2011-6-8',1500 union

select 1,'孫五','2011-10-1',11000 union
select 1,'孫五','2011-10-2',12000 union
select 1,'王五','2011-10-3',9000 union
select 1,'孫五','2011-10-4',4000 union
select 1,'張三','2011-10-1',11000 union
select 1,'張三','2011-10-3',2000 union
select 1,'張三','2011-10-4',4000 union
select 1,'李四','2011-10-6',27000 union
select 1,'趙六','2011-10-5',9000 union
select 1,'錢(qián)七','2011-10-8',3000
go
-----我們需要找出在每次的促銷(xiāo)活動(dòng)中，其銷(xiāo)售總額大于 等于
---所有其他職員銷(xiāo)售額的職員及促銷(xiāo)事件。
---說(shuō)明：謂詞a2.name<>a.name將其他職員從子查詢(xún)合計(jì)中排除出去
---------謂詞Between 中的子查詢(xún)確保我們使用了正確的促銷(xiāo)日期

--方法一：
select a.name,b.activity,SUM(a.price) as totalprice
from sales a ,Promotions as b
where a.saledate between b.sdate and b.edate
group by a.name,b.activity
having SUM(price)>= all(select SUM(price) from sales a2
where a2.name<>a.name and a2.saledate between
(
select sdate from Promotions as b2 where b2.activity=b.activity
)
and (select edate from Promotions b3
where b3.activity=b.activity)
group by a2.name)
-----------------

---方法二：
---說(shuō)明： 如果促銷(xiāo)活動(dòng)時(shí)間是不重疊的，則promotions表中只有一個(gè)主鍵列，這樣在group by
--子句中使用(activity,sdate,edate)將不會(huì)改變。但是它將使having子句可以使用sdate和edate
select a.name,b.activity,SUM(a.price) as totalprice
from sales a ,Promotions as b
where a.saledate between b.sdate and b.edate
group by b.activity,b.sdate,b.edate,a.name
having SUM(price)>= all(select SUM(price) from sales a2
where a2.name<>a.name and a2.saledate between
b.sdate
and b.edate
group by a2.name)

go

--方法三：
---使用cte(sql 2005以后的版本)
with clearksTotal(name,activity,totalprice) as
(
select a.name,b.activity,SUM(price)
from sales a ,Promotions b
where a.saledate between b.sdate and b.edate
group by a.name,b.activity
)

select c1.name,c1.activity,c1.totalprice
from clearksTotal c1
where totalprice=(select MAX(c2.totalprice) from clearksTotal c2
where c1.activity=c2.activity)
go
drop table Promotions
go
drop table sales