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内连接：把两个表中数据对应的数据查出来
外连接：以某个表为基础把对应数据查出来
首先创建数据库中的表，数据库代码如下：
/* Navicat MySQL Data Transfer Source Server : localhost_3306 Source Server Version : 50150 Source Host : localhost:3306 Source Database : store Target Server Type : MYSQL Target Server Version : 50150 File Encoding : 65001 Date: 2010-12-15 16:27:53 */ SET FOREIGN_KEY_CHECKS=0; -- ---------------------------- -- Table structure for `grade` -- ---------------------------- DROP TABLE IF EXISTS `grade`; CREATE TABLE `grade` ( `no` int(11) NOT NULL AUTO_INCREMENT, `grade` int(11) NOT NULL, PRIMARY KEY (`no`) ) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of grade -- ---------------------------- INSERT INTO grade VALUES (‘1‘, ‘90‘); INSERT INTO grade VALUES (‘2‘, ‘80‘); INSERT INTO grade VALUES (‘3‘, ‘70‘); -- ---------------------------- -- Table structure for `student` -- ---------------------------- DROP TABLE IF EXISTS `student`; CREATE TABLE `student` ( `no` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(20) NOT NULL, PRIMARY KEY (`no`) ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8; -- ---------------------------- -- Records of student -- ---------------------------- INSERT INTO student VALUES (‘1‘, ‘a‘); INSERT INTO student VALUES (‘2‘, ‘b‘); INSERT INTO student VALUES (‘3‘, ‘c‘); INSERT INTO student VALUES (‘4‘, ‘d‘);
student表中的字段分别是no和name，grade表中的字段是no和grade。两张表中的no都代表的是学生的学号。
查询student表的结果：
mysql> select * from grade; +----+-------+ | no | grade | +----+-------+ | 1 | 90 | | 2 | 80 | | 3 | 70 | +----+-------+ 3 rows in set
查询grade表的结果：
mysql> select * from student; +----+------+ | no | name | +----+------+ | 1 | a | | 2 | b | | 3 | c | | 4 | d | +----+------+ 4 rows in set
内连接 inner join（查找条件中对应的数据，no4没有数据不列出来）
mysql> select * from student s inner join grade g on s.no=g.no; +----+------+----+-------+ | no | name | no | grade | +----+------+----+-------+ | 1 | a | 1 | 90 | | 2 | b | 2 | 80 | | 3 | c | 3 | 70 | +----+------+----+-------+ 3 rows in set
左连接（左表中所有数据，右表中对应数据）
mysql> select * from student as s left join grade as g on s.no=g.no; +----+------+------+-------+ | no | name | no | grade | +----+------+------+-------+ | 1 | a | 1 | 90 | | 2 | b | 2 | 80 | | 3 | c | 3 | 70 | | 4 | d | NULL | NULL | +----+------+------+-------+ 4 rows in set
右连接（右表中所有数据，左表中对应数据）
mysql> select * from student as s right join grade as g on s.no=g.no; +----+------+----+-------+ | no | name | no | grade | +----+------+----+-------+ | 1 | a | 1 | 90 | | 2 | b | 2 | 80 | | 3 | c | 3 | 70 | +----+------+----+-------+ 3 rows in set
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原文地址：http://www.cnblogs.com/zzzzw/p/4869719.html
