Table of contents
Execution order of database sql
How to write inner and outer joins
6. Design an attendance system
7. Design a school dormitory management system
8. Design a vehicle violation system
9. Design a school canteen management system
relational table
One-to-many (1:n) creates a foreign key in the many table (the foreign key corresponds to the primary key in the one table).
Many-to-many (m:n) creates an additional intermediate table. The intermediate table has at least two fields, which serve as foreign keys pointing to the primary keys of both many-to-many parties.
Execution order of database sql
1. A sql statement usually includes:
select
from
join
where
group by
having
order by
aggregate function
limit
top
2. A brief discussion of the execution sequence:
1) First of all, make sure that it is not executed from left to right according to the order of statements we wrote.
2) Get the result set---->Specify certain fields of the query-->Sort according to certain contents
- 1. First execute from and join to determine the relationship between tables and obtain a preliminary -----> result set 1
- 2.where filter the result set 1 to get –> result set 2
- Three.Group by -->Result set 3
- 4. Filter result set 3 by having and get ----> result set 4
- Specify the fields to query:
- 5.Select specifies the field to be queried, or it can be an aggregate function -> deduplication of results
- 6. Merge the grouped result sets and sort them according to the order by conditions
- 7. If there is a limit or top , it will be executed at the end.
How to write inner and outer joins
inner join
1.A join B on A.x = B.x;
2.A inner join B on Ax = Bx; inner can be omitted
3.where A.x = B.x;
select * A join B on A.x = B.x where...
select * from A,B where A.x = B.x...
outer join
1.A left join B on A.x = B.x;
2.A right join B on A.x = B.x;
select * A left join B on A.x = B.x where...
1. Design a product list
Design a product table that contains the following fields: product name, product price, product inventory, and product description.
drop table if exists product;
create table product(
name varchar(20), --商品名称 varchar() 使用字符集UTF8 一个汉字占三个字节
price decimal(11,2), --商品价格 使用decimal() 比double精确
storage int, --商品库存
description varchar(100) --商品描述。
);2. Design a teacher list
Design a teacher table that contains the following fields: name, age, height, weight, gender, education, birthday, and ID number.
drop table if exists teacher;
create table teacher(
name varchar(20),
age int,
height double,
weight double,
sex bit,
birthday TIMESTAMP, timestamp 时间日期型 精确到秒 格式 2000-1-1 10:05:00
id_number varchar(18)
);3. Design a book list
Design a book table, including the following fields: book name, book author, book price, book category
drop table if exists book;
create table book(
name varchar(20),
author varchar(20),
price decimal(11,2),
category varchar(20)
);4. Query practice
Query in the article table for articles whose publication date create_date is from 10:30 am on January 1, 2019 to 4:02 pm on November 10, 2019.
select * from article where create_date between '2019-01-01 10:30:00' and '2019-11-10 16:02:00';
5. Query practice
In the query article table, the article title is empty, or the publication date create_date is after January 1, 2019.
select * from article where title is null or create_date > '2019-01-01 00:00:00';
6. Design an attendance system
Attendance system, including employee table, attendance record table employee and record one-to-many
-- 主要考虑记录表中的记录信息，是如何关联到员工表，员工与记录关系为1:m。
create table emp(
id int primary key,
name varchar(20)
);
create table info(
id int primary key,
emp_id int,
info_date timestamp,
foreign key (emp_id) references emp(id)
);7. Design a school dormitory management system
The school dormitory management system is required to include dormitory information, student information, and daily dormitory inspection records. One-to-many between dormitories and students, one-to-many between dormitories and ward rounds
-- 主要考虑学生与宿舍的关系：m:1，宿舍的查房记录是根据宿舍来查的，与宿舍有关系，一个宿舍可以多次查房，宿舍与查房记录是1:m的关系
create table dormitory(
id int primary key,
number varchar(20)
);
create table student(
id int primary key,
name varchar(20),
dormitory_id int,
foreign key (dormitory_id) references dormitory(id)
);
create table info(
id int primary key,
dormitory_id int,
status bit,
info_date timestamp,
foreign key (dormitory_id) references dormitory(id)
);8. Design a vehicle violation system
The vehicle violation system includes user table, vehicle table, and violation information table. The violation information table contains the violation information of the user and the vehicle.
-- 用户可以拥有多辆车，关系为1:m，
题目已经说明违章信息包括用户和车辆，说明违章信息表中要记录用户和车辆
，一个用户可以有多次违章记录，用户与违章记录关系为1:m，
一辆车也可以有多次违章记录，车辆与违章记录关系也为1:m
create table user(
id int primary key,
name varchar(20)
);
create table cars(
id int primary key,
name varchar(20),
user_id int,
foreign key (user_id) references user(id)
);
create table info(
id int primary key,
user_id int,
cars_id int,
foreign key (user_id) references user(id), --两个外键
foreign key (cars_id) references cars(id)
);9. Design a school canteen management system
The school canteen management system includes the canteen table, canteen warehouse table, and warehouse charge record table.
-- 一个食堂有多个仓口卖饭，关系为1:m，
--每个仓口卖饭可以有很多次，仓口与收费记录也是1:m
create table hall(
id int primary key,
name varchar(20)
);
create table hall_opening(
id int primary key,
name varchar(20),
hall_id int,
foreign key (hall_id) references hall(id)
);
create table info(
id int primary key,
price int,
info_date timestamp,
hall_opening_id int,
foreign key (hall_opening_id) references hall_opening(id)
);10. There is an employee table emp, with fields: name, gender, department, and salary. Query the following data:
1. Check the average salary of male and female employees
select sex,avg(salsry) as 'average salary' from emp group by sex;
2. Check the total salary of each department
select depart,sum(salsry) from emp group by depart;
3. Query the department with the second highest total salary
select depart ,sum(salary) from emp group by depart order by sum(salary) desc limit 1,1;
4. Query employee information with duplicate names
select name from emp group by name having count(name)>1;
5. Query the average salary of male employees in each department whose salary is greater than 10,000
select depart,avg(salary) from emp where sex = '男' and salary>10000 group by depart;
11. Real interview questions
There are employee tables, department tables and salary tables. Write the corresponding SQL according to the query conditions [Tongcheng Yilong 2020 School Recruitment Written Test Questions]
Now there are employee table, department table and salary table. The fields of the department table department include department_id, name; the fields of the employee table staff include staff_id, name, age, depart_id; the fields of the salary table salary include salary_id, staff_id, salary, and month.
(Problem a): Find the total department salary of each department in the month '2016-09'
select depart.name,sum(salary) from depart,staff,salary where depart.depart_id = staff.depart_id and staff.staff_id = salary.staff_id and month between '2016-09-01 00:00:00' and '2016-09-30 23:59:59' group by depart.depart_id;
select
depart.name,sum(salary)
from
depart join staff on depart.depart_id = staff.depart_id
join salary on staff.staff_id = salary.staff_id
where
month between '2016-09-01 00:00:00' and '2016-09-30 23:59:59'
group by
depart.depart_id;
(Question b): Find the number of people in each department and require the output of the department name and number of people.
select depart.depart_id,count(staff_id) from depart,staff where depart.depart_id = staff.depart_id group by depart.depart_id;
(Question c): Find the monthly salary expenditure of each department of the company, and require the output of the month and the total salary of this month.
select depart.depart_id ,mouth,sum(salary) from depart,staff,salary where depart.depart_id = staff.depart_id and staff.staff_id = salary.staff_id group by depart.depart_id ;
12.Real interview questions
Write the query conditions for the following database [SJTU Sinuo 2020 written examination questions]
There are two tables as follows:
表A(varchar(32) name， int grade)
Data: zhangshan 80, lisi 60, wangwu 84
表B(varchar(32) name， int age)
Data: zhangshan 26, lisi 24, wangwu 26, wutian 26
Write the SQL statement to get the following query results:
| NAME | GRADE | AGE |
| --------- | ----- | ---- |
| zhangshan | 80 | 26 |
| lisi | 60 | 24 |
| wangwu | 84 | 26 |
| wutian | null | 26 |
select B.name , grade,age from B left join A on B.name = A.name;