SQL练习合集

175 组合两个表 combine-two-tables

表1: Person

列名	类型
PersonId	int
FirstName	varchar
LastName	varchar

PersonId 是上表主键
表2: Address

列名	类型
AddressId	int
PersonId	int
City	varchar
State	varchar

AddressId 是上表主键
编写一个 SQL 查询，满足条件：无论 person 是否有地址信息，都需要基于上述两表提供 person 的以下信息：
FirstName, LastName, City, State

解：

SELECT FirstName, LastName, City, State FROM Person left join Address on Person.PersonId = Address.PersonId; 

176 第二高的薪水 second-highest-salary

编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。

Id	Salary
1	100
2	200
3	300

例如上述 Employee 表，SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水，那么查询应返回 null。

SecondHighestSalary
200

解:

SELECT MAX(Salary) AS SecondHighestSalary FROM Employee WHERE Salary NOT IN ( SELECT MAX(Salary) FROM Employee)S;

或

SELECT (SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1,1) AS SecondHighestSalary;

177 第N高的薪水 nth-highest-salary

编写一个 SQL 查询，获取 Employee 表中第 n 高的薪水（Salary）

Id	Salary
1	100
2	200
3	300

例如上述 Employee 表，n = 2 时，应返回第二高的薪水 200。如果不存在第 n 高的薪水，那么查询应返回 null。

getNthHighestSalary(2)
200

解:

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  SET N = N-1;
  RETURN (
      # Write your MySQL query statement below.
      SELECT (SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N,1) AS SecondHighestSalary
  );
END

178 分数排名 rank-scores

编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”

Id	Score
1	3.50
2	3.65
3	4.00
4	3.85
5	4.00
6	3.65

例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：

Score	Rank
4.00	1
4.00	1
3.85	2
3.65	3
3.65	3
3.50	4

解:

SELECT Score, (SELECT count(DISTINCT score) FROM Scores WHERE score >= s.score) AS Rank FROM Scores s ORDER BY Score DESC ;

180 连续出现的数字 consecutive-numbers

编写一个 SQL 查询，查找所有至少连续出现三次的数字。

Id	Num
1	1
2	1
3	1
4	2
5	1
6	2
7	2

例如，给定上面的 Logs 表， 1 是唯一连续出现至少三次的数字。

ConsecutiveNums
1

#select distinct l1.Num as ConsecutiveNums from Logs as l1 inner join Logs l2 on l2.Id = l1.Id + 1 and l2.Num = l1.Num inner join Logs as l3 on l3.Id = l2.Id + 1 and l3.Num = l2.Num

或

select distinct l1.Num as ConsecutiveNums from Logs as l1,Logs as l2,Logs as l3 where l2.Id = l1.Id + 1 and l2.Num = l1.Num and  l3.Id = l2.Id + 1 and l3.Num = l2.Num

181 超过经理收入的员工 employees-earning-more-than-their-managers

Employee 表包含所有员工，他们的经理也属于员工。每个员工都有一个 Id，此外还有一列对应员工的经理的 Id

Id	Name	Salary	ManagerId
1	Joe	70000	3
2	Henry	80000	4
3	Sam	60000	NULL
4	Max	90000	NULL

给定 Employee 表，编写一个 SQL 查询，该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中，Joe 是唯一一个收入超过他的经理的员工。

Employee
Joe

解:

select e1.Name as Employee from Employee as e1,Employee as e2 where e1.ManagerId = e2.Id and e1.Salary > e2.Salary

182 查找重复的电子邮件 duplicate-emails

编写一个 SQL 查询，查找 Person 表中所有重复的电子邮箱。
示例：

Id	Email
1	[email protected]
2	[email protected]
3	[email protected]

根据以上输入，你的查询应返回以下结果：

Email
[email protected]

说明：所有电子邮箱都是小写字母。

解:

select Email from Person group by Email having count(Email) > 1 

183 从不订购的客户 customers-who-never-order

某网站包含两个表，Customers 表和 Orders 表。编写一个 SQL 查询，找出所有从不订购任何东西的客户
Customers 表：

Id	Name
1	Joe
2	Henry
3	Sam
4	Max

Orders 表：

Id	CustomerId
1	3
2	1

例如给定上述表格，你的查询应返回：

Customers
Henry
Max

解:

select Name as Customers from Customers where Id not in (select CustomerId from Orders)

或

select Name as Customers from Customers left join Orders ON Customers.Id = Orders.CustomerId WHERE CustomerId IS NULL;

184 部门工资最高的员工 department-highest-salary

Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。

Id	Name	Salary	DepartmentId
1	Joe	70000	1
2	Henry	80000	2
3	Sam	60000	2
4	Max	90000	1

Department 表包含公司所有部门的信息。

Id	Name
1	IT
2	Sales

编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

Department	Employee	Salary
IT	Max	90000
Sales	Henry	80000

解:

select Department.Name as Department,Employee.Name as Employee,Salary from Employee,Department where Employee.DepartmentId = Department.Id and (Employee.Salary,Employee.DepartmentId) in (select max(Salary),DepartmentId from Employee group by DepartmentId);

185 部门工资前三高的员工 department-top-three-salaries

Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id 。

Id	Name	Salary	DepartmentId
1	Joe	70000	1
2	Henry	80000	2
3	Sam	60000	2
4	Max	90000	1
5	Janet	69000	1
6	Randy	85000	1

Department 表包含公司所有部门的信息。

Id	Name
1	IT
2	Sales

编写一个 SQL 查询，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：

Department	Employee	Salary
IT	Max	90000
IT	Randy	85000
IT	Joe	70000
Sales	Henry	80000
Sales	Sam	60000

解:

select Department.Name as Department,Employee.Name as Employee,Employee.Salary from Employee join Department on Employee.DepartmentId = Department.Id where (select count(distinct e1.Salary) from Employee e1 where e1.DepartmentId = Employee.DepartmentId and Employee.Salary < e1.Salary) < 3 order by Department.Name, Employee.Salary desc

196 删除重复的电子邮箱 delete-duplicate-emails

编写一个 SQL 查询，来删除 Person 表中所有重复的电子邮箱，重复的邮箱里只保留 Id 最小 的那个。

Id	Email
1	[email protected]
2	[email protected]
3	[email protected]

Id 是这个表的主键。
例如，在运行你的查询语句之后，上面的 Person 表应返回以下几行:

Id	Email
1	[email protected]
2	[email protected]

解:

DELETE FROM Person WHERE Id NOT IN (SELECT Id from (SELECT MIN(Id) AS id FROM Person GROUP BY Email) as MinId)

197 上升的温度 rising-temperature

给定一个 Weather 表，编写一个 SQL 查询，来查找与之前（昨天的）日期相比温度更高的所有日期的 Id。

Id(INT)	RecordDate(DATE)	Temperature(INT)
1	2015-01-01	10
2	2015-01-02	25
3	2015-01-03	20
4	2015-01-04	30

例如，根据上述给定的 Weather 表格，返回如下 Id:

Id
2
4

解:

select w1.Id from Weather as w1 join Weather as w2 where w1.RecordDate = date_add(w2.RecordDate, interval 1 day) and w1.Temperature > w2.Temperature
#可以用datediff(day,w1.RecordDate,w2.RecordDate)

595 大的国家 big-countries

这里有张 World 表

name	continent	area	population	gdp
Afghanistan	Asia	652230	25500100	20343000
Albania	Europe	28748	2831741	12960000
Algeria	Africa	2381741	37100000	188681000
Andorra	Europe	468	78115	3712000
Angola	Africa	1246700	20609294	100990000

如果一个国家的面积超过300万平方公里，或者人口超过2500万，那么这个国家就是大国家。
编写一个SQL查询，输出表中所有大国家的名称、人口和面积。
例如，根据上表，我们应该输出:

name	population	area
Afghanistan	25500100	652230
Algeria	37100000	2381741

解:

select name,population,area from World where area > 3000000 or population > 25000000 

596 超过5名学生的课 classes-more-than-5-students

有一个courses 表 ，有: student (学生) 和 class (课程)。
请列出所有超过或等于5名学生的课。
例如,表:

student	class
A	Math
B	English
C	Math
D	Biology
E	Math
F	Computer
G	Math
H	Math
I	Math

应该输出:

class
Math

解:

select class from courses group by class having count(distinct student) > 4

620 有趣的电影 not-boring-movies

某城市开了一家新的电影院，吸引了很多人过来看电影。该电影院特别注意用户体验，专门有个 LED显示板做电影推荐，上面公布着影评和相关电影描述。
作为该电影院的信息部主管，您需要编写一个 SQL查询，找出所有影片描述为非 boring (不无聊) 的并且 id 为奇数 的影片，结果请按等级 rating 排列。
例如，下表 cinema:

id	movie	description	rating
1	War	great 3D	8.9
2	Science	fiction	8.5
3	irish	boring	6.2
4	Ice song	Fantacy	8.6
5	House card	Interesting	9.1

对于上面的例子，则正确的输出是为：

id	movie	description	rating
5	House card	Interesting	9.1
1	War	great 3D	8.9

解:

select id,movie,description,rating from cinema where description != 'boring' and id%2 = 1 order by rating desc

627 交换工资 swap-salary

给定一个 salary 表，如下所示，有 m=男性 和 f=女性 的值 。交换所有的 f 和 m 值（例如，将所有 f 值更改为 m，反之亦然）。要求使用一个更新（Update）语句，并且没有中间临时表。
请注意，你必须编写一个 Update 语句，不要编写任何 Select 语句。
例如:

id	name	sex	salary
1	A	m	2500
2	B	f	1500
3	C	m	5500
4	D	f	500

运行你所编写的更新语句之后，将会得到以下表:

id	name	sex	salary
1	A	f	2500
2	B	m	1500
3	C	f	5500
4	D	m	500

解:

update salary set sex = case sex when 'm' then 'f' when 'f' then 'm' end

或

update salary set sex = if(sex = 'm','f','m')

262 行程和用户 Trips and Users

Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型，枚举成员为 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。

Id	Client_Id	Driver_Id	City_Id	Status	Request_at
1	1	10	1	completed	2013-10-01
2	2	11	1	cancelled_by_driver	2013-10-01
3	3	12	6	completed	2013-10-01
4	4	13	6	cancelled_by_client	2013-10-01
5	1	10	1	completed	2013-10-02
6	2	11	6	completed	2013-10-02
7	3	12	6	completed	2013-10-02
8	2	12	12	completed	2013-10-03
9	3	10	12	completed	2013-10-03
10	4	13	12	cancelled_by_driver	2013-10-03

Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止，Role 则是一个表示（‘client’, ‘driver’, ‘partner’）的枚举类型。

Users_Id	Banned	Role
1	No	client
2	Yes	client
3	No	client
4	No	client
10	No	driver
11	No	driver
12	No	driver
13	No	driver

写一段 SQL 语句查出 2013年10月1日 至 2013年10月3日 期间非禁止用户的取消率。基于上表，你的 SQL 语句应返回如下结果，取消率（Cancellation Rate）保留两位小数。

Day	Cancellation Rate
2013-10-01	0.33
2013-10-02	0.00
2013-10-03	0.50

解:

select request_at as Day, 
round((sum(case when status = 'cancelled_by_driver' or status = 'cancelled_by_client' then 1 else 0 end)) / count(status) ,2) as 'Cancellation Rate'
from Trips where
Client_Id in (select Users_Id from Users where Banned = 'No')
and Request_at >= '2013-10-01' and Request_at <= '2013-10-03'
group by request_at

626 换座位 Exchange Seats

小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。
其中纵列的 id 是连续递增的
小美想改变相邻俩学生的座位。
你能不能帮她写一个 SQL query 来输出小美想要的结果呢？
示例：

id	student
1	Abbot
2	Doris
3	Emerson
4	Green
5	Jeames

假如数据输入的是上表，则输出结果如下：

id	student
1	Doris
2	Abbot
3	Green
4	Emerson
5	Jeames

注意：
如果学生人数是奇数，则不需要改变最后一个同学的座位。

解:

select (case 
           when id%2=1 and id!=(select count(*) from seat) then id+1 
           when id%2=0 then id-1 
           else id
           end) id, student
from seat
order by id asc

601 体育馆的人流量 Human Traffic of Stadium

X 市建了一个新的体育馆，每日人流量信息被记录在这三列信息中：序号 (id)、日期 (date)、 人流量 (people)。

请编写一个查询语句，找出高峰期时段，要求连续三天及以上，并且每天人流量均不少于100。

例如，表 stadium：

id	date	people
1	2017-01-01	10
2	2017-01-02	109
3	2017-01-03	150
4	2017-01-04	99
5	2017-01-05	145
6	2017-01-06	1455
7	2017-01-07	199
8	2017-01-08	188

对于上面的示例数据，输出为：

id	date	people
5	2017-01-05	145
6	2017-01-06	1455
7	2017-01-07	199
8	2017-01-08	188

Note:
每天只有一行记录，日期随着 id 的增加而增加。

解:

select distinct s4.id,s4.visit_date,s4.people from stadium s1,stadium s2,stadium s3,stadium s4
where s1.id = s2.id + 1 
and s2.id = s3.id + 1
and s1.people >= 100
and s2.people >= 100
and s3.people >= 100
and s4.id in (s1.id,s2.id,s3.id)

书上习题

create table Student(
 s_no char(6) primary key,
 class_no char(6) not null,
 s_name varchar(10) not null,
 s_sex char(2) check(s_sex in ('男','女')),
 s_birthday datetime
);

create table Class(
class_no char(6) primary key,
class_name char(20) not null,
class_special varchar(20),
class_dept char(20)
);

create table Course(
    course_no char(5) primary key,
    course_name char(20) not null,
    course_score numeric(6,2)
);

create table Choise(
    s_no char(6),
    course_no char(5),
    score numeric(6,1)
);

create table Teacher(
    t_no char(6) primary key,
    t_name varchar(10) not null,
    t_sex char(2) check(s_sex in ('男','女')),
    t_birthday datetime,
    t_title char(10)
);

create table Teaching(
    course_no char(5),
    t_no char(6)
);

insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991101','js9901','张彬','男','1981-10-1');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991102','js9901','王蕾','女','1980-8-8');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991103','js9901','李建国','男','1981-4-5');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991104','js9901','李平方','男','1981-5-12');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991201','js9902','陈东辉','男','1980-2-8');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991202','js9902','葛鹏','男','1979-12-23');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991203','js9902','潘桃枝','女','1980-2-6');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('991204','js9902','姚一峰','男','1981-5-7');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('001101','js0001','宋大芳','男','1980-4-9');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('001102','js0001','许辉','女','1978-8-1');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('001201','js0002','王一山','男','1980-12-4');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('001202','js0002','牛莉','女','1981-6-9');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('002101','xx0001','李丽丽','女','1981-9-19');
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values('002102','xx0001','李王','男','1980-9-23');

insert into Class(class_no,class_name,class_special,class_dept) values('js9901','计算机99-1','计算机','计算机系');
insert into Class(class_no,class_name,class_special,class_dept) values('js9902','计算机99-2','计算机','计算机系');
insert into Class(class_no,class_name,class_special,class_dept) values('js0001','计算机00-1','计算机','计算机系');
insert into Class(class_no,class_name,class_special,class_dept) values('js0002','计算机00-2','计算机','计算机系');
insert into Class(class_no,class_name,class_special,class_dept) values('xx0001','信息00-1','信息','信息系');
insert into Class(class_no,class_name,class_special,class_dept) values('xx0002','信息00-2','信息','信息系');

insert into Course(course_no,course_name,course_score) values('01001','计算机基础',3);
insert into Course(course_no,course_name,course_score) values('01002','程序设计语言',5);
insert into Course(course_no,course_name,course_score) values('01003','数据结构',6);
insert into Course(course_no,course_name,course_score) values('02001','数据库原理与应用',6);
insert into Course(course_no,course_name,course_score) values('02002','计算机网络',6);
insert into Course(course_no,course_name,course_score) values('02003','微机原理与应用',8);

insert into Choise(s_no,course_no,score) values('991101','01001',88.0);
insert into Choise(s_no,course_no,score) values('991102','01001',);
insert into Choise(s_no,course_no) values('991103','01001',91.0);
insert into Choise(s_no,course_no,score) values('991104','01001',78.0);
insert into Choise(s_no,course_no,score) values('991201','01001',67.0);
insert into Choise(s_no,course_no,score) values('991101','01002',90.0);
insert into Choise(s_no,course_no,score) values('991102','01002',58.0);
insert into Choise(s_no,course_no,score) values('991103','01002',71.0);
insert into Choise(s_no,course_no,score) values('991104','01002',85.0);

insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000001','李英','女','1964-11-3','讲师');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000002','王大山','男','1955-3-7','副教授');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000003','张朋','男','1960-10-5','讲师');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000004','陈为军','男','1970-3-2','助教');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000005','宋浩然','男','1966-12-4','讲师');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000006','徐红霞','女','1951-5-8','副教授');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000007','徐永军','男','1948-4-8','教授');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000008','李桂清','女','1940-11-3','教授');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000009','王一帆','女','1962-5-9','讲师');
insert into Teacher(t_no,t_name,t_sex,t_birthday,t_title) values('000010','田丰','男','1972-11-5','助教');

insert into Teaching(course_no,t_no) values('01001','000001');
insert into Teaching(course_no,t_no) values('01002','000002');
insert into Teaching(course_no,t_no) values('01003','000002');
insert into Teaching(course_no,t_no) values('02001','000003');
insert into Teaching(course_no,t_no) values('02002','000004');
insert into Teaching(course_no,t_no) values('01001','000005');
insert into Teaching(course_no,t_no) values('01002','000006');
insert into Teaching(course_no,t_no) values('01003','000007');
insert into Teaching(course_no,t_no) values('02001','000007');
insert into Teaching(course_no,t_no) values('02002','000008');

#查询所有学生的基本信息
select s_no,class_no,s_name,s_sex,s_birthday from Student;
#查询所有学生　显示学号姓名
select s_no,s_name from Student;
#查询男同学　显示学号姓名出生日期
select s_no,s_name,s_birthday from Student where s_sex = '男';
＃查询出生日期在80-1-1前女学生
select s_no,class_no,s_name,s_sex,s_birthday from Student where s_sex='女' and s_birthday<'1980-01-01';
#查询李姓男学生
select s_no,class_no,s_name,s_sex,s_birthday from Student where s_sex='男' and s_name like "李%"; 
#查询名字含一
select s_no,class_no,s_name,s_sex,s_birthday from Student where s_name like "%一%";
#查询职称不是讲师
select t_no,t_name,t_sex,t_birthday,t_title from Teacher where t_title != '讲师';
#查询未参加考试的学生
select s_no,course_no,score from Choise where score is NULL;
#查询考试不及格的学生并按成绩排序
select s_no,course_no,score from Choise where score < 60 ORDER BY score ASC;
#查询课程号为01001 02001 02003所有课程
select course_no,course_name,course_score from Course where course_no in ('01001','02001','02003');
#查询1970年出生的讲师
select t_no,t_name,t_sex,t_birthday,t_title from Teacher where t_birthday < '1970';
#查询各课程号选修人数
select course_no,count(course_no) from Choise group by course_no ;
#查询教授两门课程以上的讲师
select t_no,count(course_no) from Teaching group by t_no having count(course_no)=>2  ;
#查询01001平均分最低分最高分
select avg(score),min(score),max(score) from Choise where course_no = '01001';
#查询1960年后出生职称为讲师并按出生日期升序
select t_no,t_name,t_sex,t_birthday,t_title from Teacher where t_birthday > '1960' and t_title = '讲师'　ORDER BY t_birthday ASC;
#查询所有学生选课和成绩s_no s_name course_no score
select Student.s_no,s_name,course_no,score from Student join Choise on Student.s_no = Choise.s_no;
#查询计算机99-1同学的选课及成绩s_no s_name course_no course_name score
select Student.s_no,s_name,Course.course_no,course_name,score from Student join Choise,Course,Class where Student.s_no = Choise.s_no and Course.course_no = Choise.course_no and Student.class_no = Class.class_no and Class.class_name = '计算机99-1'; 
#查询所有学生及格科目
select Student.s_no,s_name,Course.course_no,course_name,score from Student join Choise,Course where Student.s_no = Choise.s_no and Course.course_no = Choise.course_no and Choise.score >= 60; 
#查询所有学生平均成绩和所选科目数
select Student.s_no,s_name,avg(score),count(score) from Choise join Student  where Student.s_no = Choise.s_no group by s_no;
#查询未参加考试的学生和课程
select Student.s_no,s_name,Course.course_no,course_name,score from Choise join Student,Course where Course.course_no = Choise.course_no and Student.s_no = Choise.s_no and score is NULL;
#查询不及格的学生和课程
select Student.s_no,s_name,Course.course_no,course_name,score from Choise join Student,Course where Course.course_no = Choise.course_no and Student.s_no = Choise.s_no and score < 60;
#查询"程序设计语言"的所有同学和成绩　ＡＮＹ
select s_name,score from Student,Choise where course_no = any(select course_no from Course where course_name = '程序设计语言') and Student.s_no = Choise.s_no;
#查询所有教师的任课
select t_name,course_name from Teaching,Teacher,Course where Teacher.t_no = Teaching.t_no and Teaching.course_no = Course.course_no;
#查询李建国的同学　子查询
select s_name from Student where class_no = (select class_no from Student where s_name = '李建国') and s_name != '李建国';
#查询没有选修计算机基础ＮＯＴＥＸＩＳＴＳ
select s_name from Student,Choise where NOT EXISTS (select course_no from Course where course_name = '计算机基础' and course_no = Choise.course_no) and Student.s_no = Choise.s_no;
#查询主讲数据库原理和应用和主讲数据结构的教师姓名　ＵＮＩＯＮ
select t_name from Teacher,Teaching,Course where Teacher.t_no = Teaching.t_no and Teaching.course_no = Course.course_no and Course.course_name = '数据库原理与应用' UNION select t_name from Teacher,Teaching,Course where Teacher.t_no = Teaching.t_no and Teaching.course_no = Course.course_no and Course.course_name = '数据结构'; 
#插入学生的存储过程
delimiter $$
create procedure insertStudent(
 in s_no char(6),
 in class_no char(6),
 in s_name varchar(10),
 in s_sex char(2),
 in s_birthday datetime
)
begin
insert into Student(s_no,class_no,s_name,s_sex,s_birthday) values(s_no,class_no,s_name,s_sex,s_birthday);
end$$
delimiter ;
#插入课程　学分默认2
#delimiter $$
#create procedure insertCourse(
#in course_no char(5),
#in course_name char(20),
#in course_score numeric(6,2) DEFAULT 2
#)
#begin
#insert into Course(course_no,course_name,course_score) values(course_no,course_name,course_score);
#end$$
#delimiter ;
#根据姓名查询学生信息
delimiter $$
create procedure queryStudent(
 in name varchar(10)
)
begin
select s_no,s_name from Student where s_name = name;
end$$
delimiter ;
#触发器
drop trigger StudentTrigger;
delimiter $$
create trigger StudentTrigger
AFTER INSERT ON Student FOR EACH ROW
begin
select NEW.s_no into @arg1;
select NEW.s_name into @arg2;
end$$
delimiter ;