我对多表关联的总结://以下若有理解错误请指出,谢谢! 外连接: RIGHT JOIN:以右表为基础,显示右表中的所有列,不管是否与关联条件相匹配,而左表中的数据只显示与关联条件相匹配的列,不匹配的列以NULL字符填充. FULL JOIN?:显示多个表中的所有的列,不匹配关联条件的列以NULL字符填充. 下面是我写的关于sql 多表关联的sql语句,大家可以在SQL SERVER中操作下试下,是否与上面的对于sql关联的总结相同,查询结果我就不结了,大家试下就知道了! create database mydb; use mydb; create table customers create table books create table orders insert into books(book_name,price) values('php',78); insert into orders(isbn,cust_id,order_date) values(700001,1,getdate()); select a.order_id,a.isbn,b.book_name,b.price,c.cust_name,c.cust_tel,c.cust_address,a.order_date ? 左表右表指哪些表? 第二种认为: CSDN SQL专区提问,竟请您的加盟:http://community.csdn.net/Expert/topic/5696/5696608.xml?temp=.4501764 还请牛人指点! 最终答案: -----------------[以下为网上的一点资料]------------------?
内连接??????
INNER JOIN:只显示多表之间与关联条件相匹配的列.
LEFT JOIN?:以左表为基础,显示左表中的所有列,不管是否与关联条件相匹配,而右表中的数据只显示与关联条件相匹配的列,不匹配的列以NULL字符填充.
(
cust_id int not null identity(1,1) primary key,
cust_name varchar(100) not null,
cust_tel varchar(50) not null,
cust_address varchar(100) not null?
)
(
isbn int not null identity(700001,1),
book_name varchar(100) not null,
price money not null
)
(
order_id int not null identity(1001,1),
isbn int not null,
cust_id int not null,
order_date datetime not null
)
insert into customers(cust_name,cust_tel,cust_address) values('书店a','0553-2451466','江苏省');
insert into customers(cust_name,cust_tel,cust_address) values('书店b','0553-2215266','江苏省');
insert into customers(cust_name,cust_tel,cust_address) values('书店c','0553-8754466','江苏省');
insert into customers(cust_name,cust_tel,cust_address) values('书店d','0553-4563466','江苏省');
insert into customers(cust_name,cust_tel,cust_address) values('书店e','0553-2145212','江苏省');
insert into books(book_name,price) values('java',72);
insert into books(book_name,price) values('c++',72);
insert into books(book_name,price) values('C语言',56);
insert into books(book_name,price) values('sql',53);
insert into books(book_name,price) values('asp.net',69);
insert into books(book_name,price) values('asp',43);
insert into orders(isbn,cust_id,order_date) values(700002,1,getdate());
insert into orders(isbn,cust_id,order_date) values(700003,2,getdate());
insert into orders(isbn,cust_id,order_date) values(700001,1,getdate());
insert into orders(isbn,cust_id,order_date) values(700003,1,getdate());
insert into orders(isbn,cust_id,order_date) values(700006,1,getdate());
insert into orders(isbn,cust_id,order_date) values(700001,2,getdate());
insert into orders(isbn,cust_id,order_date) values(700001,2,getdate());
insert into orders(isbn,cust_id,order_date) values(700002,2,getdate());
insert into orders(isbn,cust_id,order_date) values(700003,1,getdate());
insert into orders(isbn,cust_id,order_date) values(700003,9,getdate());--我下面是故意的,让大家看到效果
insert into orders(isbn,cust_id,order_date) values(7000025,9,getdate());
from orders as a inner join books as b?
on
a.isbn=b.isbn inner join customers as c
on
a.cust_id=c.cust_id;
select a.order_id,a.isbn,b.book_name,b.price,c.cust_name,c.cust_tel,c.cust_address,a.order_date
from orders as a left join books as b?
on
a.isbn=b.isbn left join customers as c
on
a.cust_id=c.cust_id;
select a.order_id,a.isbn,b.book_name,b.price,c.cust_name,c.cust_tel,c.cust_address,a.order_date
from orders as a right join books as b?
on
a.isbn=b.isbn right join customers as c
on
a.cust_id=c.cust_id;
select * from?
table_1 as a left join table_2 as c
on
a.id=b.id?
leftjoin table_3 as c
on
b.isbn=c.isbn
疑问中............
第一种认为:
认为最左边的表左表 table_1
右表就是除了左表的其它表,而不是最右边的表
所以这里的左表是 table_1,那右表是table_2,table_3
table_1 是table_2的左表,table_2是table_3的左表
table_1 左连接 table_2 以table_1为基础进行连接运算,得到新表即临时表(#table)。
#table???? 左连接 table_3 以table# 为基础进行连接运算,得到最终结果。
与我的总结一致!
以下是来自网上的关于关联说的比较好的文章(易懂)
来自:http://www.cnblogs.com/temptation/archive/2006/05/23/407328.aspx
表A记录如下:
aID????????????????? aNum
1???????????????????? a20050111
2???????????????????? a20050112
3???????????????????? a20050113
4???????????????????? a20050114
5???????????????????? a20050115
表B记录如下:
bID????????????????? bName
1????????????????????? 2006032401
2???????????????????? 2006032402
3???????????????????? 2006032403
4???????????????????? 2006032404
8???????????????????? 2006032408
实验如下:
1.????????????? left join
sql语句如下:?SELECT?*?FROM?A
LEFT?JOIN?B?
ON?A.aID?=?B.bID
结果如下:
aID????????????????? aNum???????????????????????????? bID???????????????????? bName
1????????????????????? a20050111?????????????????? 1???????????????????????? 2006032401
2????????????????????? a20050112?????????????????? 2??????????????????????? 2006032402
3????????????????????? a20050113?????????????????? 3??????????????????????? 2006032403
4????????????????????? a20050114?????????????????? 4??????????????????????? 2006032404
5????????????????????? a20050115?????????????????? NULL???????????????? NULL
(所影响的行数为 5 行)
结果说明:
????????????????? left join是以A表的记录为基础的,A可以看成左表,B可以看成右表,left join是以左表为准的.
换句话说,左表(A)的记录将会全部表示出来,而右表(B)只会显示符合搜索条件的记录(例子中为: A.aID = B.bID).
B表记录不足的地方均为NULL.
2.????????????? right join
sql语句如下:?SELECT?*?FROM?A
RIGHT?JOIN?B?
ON?A.aID?=?B.bID
结果如下:
aID????????????????? aNum???????????????????????????? bID???????????????????? bName
1????????????????????? a20050111?????????????????? 1???????????????????????? 2006032401
2????????????????????? a20050112?????????????????? 2??????????????????????? 2006032402
3????????????????????? a20050113?????????????????? 3??????????????????????? 2006032403
4????????????????????? a20050114?????????????????? 4??????????????????????? 2006032404
NULL????????????? NULL????????????????????????? 8??????????????????????? 2006032408
(所影响的行数为 5 行)
结果说明:
????????????????? 仔细观察一下,就会发现,和left join的结果刚好相反,这次是以右表(B)为基础的,A表不足的地方用NULL填充.
3.inner join
sql语句如下:?SELECT?*?FROM?A
INNERJOIN B?
ON?A.aID?=?B.bID
结果如下:
aID????????????????? aNum???????????????????????????? bID???????????????????? bName
1????????????????????? a20050111?????????????????? 1???????????????????????? 2006032401
2????????????????????? a20050112?????????????????? 2??????????????????????? 2006032402
3????????????????????? a20050113?????????????????? 3??????????????????????? 2006032403
4????????????????????? a20050114?????????????????? 4??????????????????????? 2006032404
结果说明:
????????????????? 很明显,这里只显示出了 A.aID = B.bID的记录.这说明inner join并不以谁为基础,它只显示符合条件的记录.
LEFT JOIN操作用于在任何的 FROM 子句中,组合来源表的记录。使用 LEFT JOIN 运算来创建一个左边外部联接。左边外部联接将包含了从第一个(左边)开始的两个表中的全部记录,即使在第二个(右边)表中并没有相符值的记录。?
语法:
FROM table1 LEFT JOIN table2 ON table1.field1 compopr table2.field2?
说明:
① table1, table2参数用于指定要将记录组合的表的名称。
② field1, field2参数指定被联接的字段的名称。且这些字段必须有相同的数据类型及包含相同类型的数据,但它们不需要有相同的名称。
③ compopr参数指定关系比较运算符:"=", "<", ">", "<=", ">=" 或 "<>"。
④ 如果在INNER JOIN操作中要联接包含Memo 数据类型或 OLE Object 数据类型数据的字段,将会发生错误。
详细解决方案
SQL多表联系关系! inner join,left join ,right join
热度:131 发布时间:2016-05-05 14:58:54.0
SQL多表关联! inner join,left join ,right join
2007-08-06 20:35
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