新建表
mysql> create table department(
id int,
name varchar(20)
);
mysql> create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
);
插入数据
mysql> insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营');
mysql> insert into employee(name,sex,age,dep_id) values
('egon','male',18,200),
('alex','female',48,201),
('wupeiqi','male',38,201),
('yuanhao','female',28,202),
('liwenzhou','male',18,200),
('jingliyang','female',18,204)
;
查看表结构和数据
mysql> desc department;
mysql> desc employee;
mysql> select * from department;
mysql> select * from employee;
二、多表连接查询
重点！！！外链接语法
ELECT 字段列表 FROM 表1 INNER|LEFT|RIGHT JOIN 表2 ON 表1.字段 = 表2.字段;
1、交叉连接：不适用任何匹配条件。生成笛卡尔积
mysql> select * from employee,department;
2、内连接：只连接匹配的行
找两张表共有的部分，相当于利用条件从笛卡尔积结果中筛选出了正确的结果
select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id;
上述sql等同于
select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
注意：department没有204这个部门，因而employee表中关于204这条员工信息没有匹配出来
3、外链接之左连接：优先显示左表全部记录
以左表为准，即找出所有员工信息，当然包括没有部门的员工；本质就是：在内连接的基础上增加左边有右边没有的结果
select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
4、外链接之右连接：优先显示右表全部记录
以右表为准，即找出所有部门信息，包括没有员工的部门；本质就是：在内连接的基础上增加右边有左边没有的结果
select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
5、全外连接：显示左右两个表全部记录
在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
注意：mysql不支持全外连接 full JOIN
强调：mysql可以使用此种方式间接实现全外连接
select * from employee left join department on employee.dep_id = department.id union select * from employee right join department on employee.dep_id = department.id;
注意 union与union all的区别：union会去掉相同的纪录
三、符合条件连接查询
示例1：以内连接的方式查询employee和department表，并且employee表中的age字段值必须大于25,即找出年龄大于25岁的员工以及员工所在的部门
select employee.name,department.name from employee inner join department on employee.dep_id = department.id where age > 25;
示例2：以内连接的方式查询employee和department表，并且以age字段的升序方式显示
select employee.id,employee.name,employee.age,department.name from employee,department where employee.dep_id = department.id and age > 25 order by age asc;
四 子查询
1）子查询是将一个查询语句嵌套在另一个查询语句中。
2）内层查询语句的查询结果，可以为外层查询语句提供查询条件。
3）子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
4）还可以包含比较运算符：= 、 !=、> 、<等
1、带IN关键字的子查询
查询平均年龄在25岁以上的部门名
select id,name from department where id in (select dep_id from employee group by dep_id having avg(age) > 25);
查看技术部员工姓名
select name from employee where dep_id in (select id from department where name='技术');
查看不足1人的部门名(子查询得到的是有人的部门id)
select name from department where id not in (select distinct dep_id from employee);
2、带比较运算符的子查询
比较运算符：=、!=、>、>=、<、<=、<>
查询大于所有人平均年龄的员工名与年龄
select name,age from emp where age > (select avg(age) from emp);
查询大于部门内平均年龄的员工名、年龄
select t1.name,t1.age from emp t1 inner join (select dep_id,avg(age) avg_age from emp group by dep_id) t2 on t1.dep_id = t2.dep_id where t1.age > t2.avg_age;
3、带EXISTS关键字的子查询
EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录。而是返回一个真假值。True或False，当返回True时，外层查询语句将进行查询；当返回值为False时，外层查询语句不进行查询
department表中存在dept_id=203，Ture
select * from employee where exists (select id from department where id=200);
department表中存在dept_id=205，False
select * from employee where exists (select id from department where id=204);
4、练习：查询每个部门最新入职的那位员工
准备表：
mysql>create table employee(
id int not null unique auto_increment,
name varchar(20) not null,
sex enum('male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date not null,
post varchar(50),
post_comment varchar(100),
salary double(15,2),
office int, #一个部门一个屋子
depart_id int
);
mysql>insert into employee(name,sex,age,hire_date,post,salary,office,depart_id) values
('egon','male',18,'20170301','老男孩驻沙河办事处外交大使',7300.33,401,1), #以下是教学部
('alex','male',78,'20150302','teacher',1000000.31,401,1),
('wupeiqi','male',81,'20130305','teacher',8300,401,1),
('yuanhao','male',73,'20140701','teacher',3500,401,1),
('liwenzhou','male',28,'20121101','teacher',2100,401,1),
('jingliyang','female',18,'20110211','teacher',9000,401,1),
('jinxin','male',18,'19000301','teacher',30000,401,1),
('成龙','male',48,'20101111','teacher',10000,401,1),
('歪歪','female',48,'20150311','sale',3000.13,402,2),#以下是销售部门
('丫丫','female',38,'20101101','sale',2000.35,402,2),
('丁丁','female',18,'20110312','sale',1000.37,402,2),
('星星','female',18,'20160513','sale',3000.29,402,2),
('格格','female',28,'20170127','sale',4000.33,402,2),
('张野','male',28,'20160311','operation',10000.13,403,3), #以下是运营部门
('程咬金','male',18,'19970312','operation',20000,403,3),
('程咬银','female',18,'20130311','operation',19000,403,3),
('程咬铜','male',18,'20150411','operation',18000,403,3),
('程咬铁','female',18,'20140512','operation',17000,403,3)
;
答案1：链表
SELECT * FROM employee AS t1 INNER JOIN (
SELECT post,max(hire_date) max_date FROM employee GROUP BY post ) AS t2 ON t1.post = t2.post
WHERE t1.hire_date = t2.max_date;
答案2：子查询
select t3.name,t3.post,t3.hire_date from employee as t3 where id in (select (select id from employee as t2 where t2.post=t1.post order by hire_date desc limit 1) from employee as t1 group by post);
五、综合练习
1、准备表、记录
mysql> create database db1;
mysql> use db1;
mysql> source /root/init.sql （从init.sql文件中导入数据）
2、题目
1)查询所有的课程的名称以及对应的任课老师姓名
select
teacher.tname,
course.cname
from
teacher
inner join course on teacher.tid = course.teacher_id;
2)查询学生表中男女生各有多少人
select gender,count(sid) from student group by gender;
3)查询物理成绩等于100的学生的姓名
select
student.sname
from
student
where
sid in (
select
student_id
from
score
inner join course on score.course_id = course.cid
WHERE
course.cname = '物理' AND score.num = 100);
4)查询平均成绩大于八十分的同学的姓名和平均成绩
select
student.sname,
t1.avg_num
from
student
inner join (
select
student_id
avg(num) AS avg_num
from
score
group by
student_id
having
avg(num) > 80
)AS t1 on student.sid=t1.student_id;
5)查询所有学生的学号，姓名，选课数，总成绩
SELECT
student.sid,
student.sname,
t1.course_num,
t1.total_num
FROM
student
LEFT JOIN (
SELECT
student_id,
COUNT(course_id) course_num,
sum(num) total_num
FROM
score
GROUP BY
student_id
) AS t1 ON student.sid = t1.student_id;
6)查询姓李老师的个数
select count(tid) from teacher where tname LIKE '李%';
7)查询没有报李平老师课的学生姓名
select
student.sname
from
student
8)查询物理课程比生物课程高的学生的学号
9）查询没有同时选修物理课程和体育课程的学生姓名
10）查询挂科超过两门(包括两门)的学生姓名和班级
11）查询选修了所有课程的学生姓名
12）查询李平老师教的课程的所有成绩记录
13）查询全部学生都选修了的课程号和课程名
14）查询每门课程被选修的次数
15）查询之选修了一门课程的学生姓名和学号
16）查询所有学生考出的成绩并按从高到低排序（成绩去重）
17）查询平均成绩大于85的学生姓名和平均成绩
18）查询生物成绩不及格的学生姓名和对应生物分数
19）查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名
20）查询每门课程成绩最好的前两名学生姓名
21）查询不同课程但成绩相同的学号，课程号，成绩
22）查询没学过“叶平”老师课程的学生姓名以及选修的课程名称；
23)查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名；
24)任课最多的老师中学生单科成绩最高的学生姓名
原文地址：http://blog.51cto.com/10630401/2110961
