关于数据库中使用 left join on ...and ...和 left join on ....where ...区别,和使用group by 要注意的情况

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