create table UserAccount
(
UID int,
UName varchar(50),
JoinDate datetime
)
insert into UserAccount values(1,'Jimmy','2006-10-3')
insert into UserAccount values(2,'Jack','2006-10-25')
insert into UserAccount values(3,'Mike','2006-11-13')
insert into UserAccount values(4,'Tom','2006-9-5')
create table ChargeHistory(UId int,ChargeDate datetime,Amount int)
insert into ChargeHistory values(1,'2007-3-1',50)
insert into ChargeHistory values(1,'2007-2-15', 30)
insert into ChargeHistory values(2,'2007-2-1' ,40)
insert into ChargeHistory values(3,'2007-1-15' ,60)
insert into ChargeHistory values(2,'2007-1-15' ,50)
insert into ChargeHistory values(3,'2007-1-1' ,100)
insert into ChargeHistory values(1,'2006-12-20' ,60)
select * from UserAccount
select * from ChargeHistory
1
Jimmy 2006-10-03 00:00:00.000 2
Jack 2006-10-25 00:00:00.000 3
Mike 2006-11-13 00:00:00.000 4
Tom 2006-09-05 00:00:00.000
1
2007-03-01 00:00:00.000 50 1
2007-02-15 00:00:00.000 30 2
2007-02-01 00:00:00.000 40 3
2007-01-15 00:00:00.000 60 2
2007-01-15 00:00:00.000 50 3
2007-01-01 00:00:00.000 100 1
2006-12-20 00:00:00.000 60
假设有一个游戏帐户充值网站
系统内有两个表
UserAccount
|
UID |
UName |
JoinDate |
|
1 |
Jimmy |
2006-10-3 |
|
2 |
Jack |
2006-10-25 |
|
3 |
Mike |
2006-11-13 |
|
4 |
Tom |
2006-9-5 |
ChargeHistory
|
UId |
ChargeDate |
Amount |
|
1 |
2007-3-1 |
50 |
|
1 |
2007-2-15 |
30 |
|
2 |
2007-2-1 |
40 |
|
3 |
2007-1-15 |
60 |
|
2 |
2007-1-15 |
50 |
|
3 |
2007-1-1 |
100 |
|
1 |
2006-12-20 |
60 |
1. 建表并填充数据
2. 完成如下操作:
A. 查询用户名为mike的充值记录,包含如下列
UName,UId,JoinDate,ChargeDate,Amount
B. 查询用户Id为1共充值多少金额.包含
UName,UId,JoinDate,Amount
C. 统计每个用户的充值总额,结果如下
|
UId |
UName |
Amount |
JoinDate |
|
1 |
Jimmy |
140 |
2006-10-3 |
|
2 |
Jack |
90 |
2006-10-25 |
|
3 |
Mike |
160 |
2006-11-13 |
|
4 |
Tom |
Null |
2006-9-5 |
select A.UID,A.UName,A.JoinDate,B.ChargeDate,B.Amount from UserAccount A
left join ChargeHistory B on A.UID=B.UID and a.uname='mike' 这里的结果是:不是我们想要的结果因为:
--通过以下结果我们可以看出:左连接是left join 嘛,无非是左边表为基础, 扫描右边表匹配的记录
--按条件 a.uname='mike', 来扫描右边表的记录
--对于右边表的每条记录, 显然 a.uname='mike' 这个条件都是成立.
Jimmy 2006-10-03 00:00:00.000 NULL NULL 2
Jack 2006-10-25 00:00:00.000 NULL NULL 3
Mike 2006-11-13 00:00:00.000 2007-01-15 00:00:00.000 60 3
Mike 2006-11-13 00:00:00.000 2007-01-01 00:00:00.000 100 4
Tom 2006-09-05 00:00:00.000 NULL NULL
select A.UID,A.UName,A.JoinDate,B.ChargeDate,B.Amount from UserAccount A
left join ChargeHistory B on A.UID=B.UID where A.UNAME='Mike'所以这里要使用where .where是left join 后筛选结果.
结果:
Mike 2006-11-13 00:00:00.000 2007-01-15 00:00:00.000 60 3
Mike 2006-11-13 00:00:00.000 2007-01-01 00:00:00.000 100 4
--但是inner join 是选择符合条件的数据出来
select A.UID,A.UName,A.JoinDate,B.ChargeDate,B.Amount from UserAccount A
inner join ChargeHistory B on A.UID=B.UID and a.uname='mike'
--group by 的操作:
--有函数存在 :sum,count,min,avg,max
--slecect多少个字段,group by 多少个字段
select A.UID,A.UName,A.JoinDate,sum(B.amount) from UserAccount A left join ChargeHistory B ON A.uID=B.UID WHERE A.UID=1
group by a.uid,A.UName,A.JoinDate
--这里还要注意 having的用法:是用来在group by 后再筛选数据的
select A.UID,A.UName,A.JoinDate,sum(B.amount) from UserAccount A left join ChargeHistory B ON A.uID=B.UID
group by a.uid,A.UName,A.JoinDate
having A.UID=1
