Detailed examples of MySQL database design issues
This article mainly shares with you an article about the design issues of MySQL database. It has a good reference value and I hope it will be helpful to everyone. Let’s follow the editor to take a look, I hope it can help everyone.
First question: Hierarchical database design
Question description: There are now about 100,000 pieces of data, recording employees in a department. Under most departments is a hierarchical structure with many sub-departments. For example, the first-level part A, the second-level departments A', B', C', and the third-level departments A", B", C". How to design the database, we need to count all the people under the second-level part A'.
Analysis:
A hierarchical database design is used here
CREATE TABLE DEPARTMENT(
DEP_ID INT UNSIGNED AUTO_INCREMENT,
DEP_NAME VARCHAR(10) NOT NULL,
PARENT_ID INT,
PRIMARY KEY(DEP_ID)
)CHARSET=utf8;
Insert data
单个插入
INSERT INTO DEPARTMENT (DEP_NAME, PARENT_ID)
VALUES
('A',NULL);
或者批量插入
INSERT INTO department VALUES(1,'A',NULL),(2,'B',1),(3,'C',1),
(4,'D',2),(5,'E',2),(6,'F',3),(7,'G',3);| #dep_name | parent_id | |
|---|---|---|
| A | NULL | |
| B | 1 | |
| C | 1 | |
| D | 2 | |
| E | 2 | |
| F | 3 | ##7 |
| 3 |
dep_id
of this level, find its
parent_id , and then connect through the left connection to get the current department and its parent department
select d1.dep_name as level1, d2.dep_name as level2, d3.dep_name as level3, d4.dep_name as level4 from department as d1 left join department as d2 on d2.parent_id = d1.dep_id left join department as d3 on d3.parent_id = d2.dep_id left join department as d4 on d4.parent_id = d3.dep_id where d1.dep_name='A';
Create a table and store department personnel information
create table people(
id INT UNSIGNED AUTO_INCREMENT,
name varchar(10) not null,
dep_id INT UNSIGNED,
departname varchar(10),
FOREIGN KEY (dep_id) REFERENCES department(dep_id),
primary key(id)
)charset=utf8;Insert relevant test data
INSERT INTO people VALUES(1,'hgy',4,'D'),(2,'abc',5,'E'),(3,'def',6,'F'),
(4,'ddd',2,'B'),(5,'eee',2,'B');
Find secondary level. The person whose department is B, and his superior department information is listed
select p.id, p.name, d1.dep_name as level1, d2.dep_name as level2, d3.dep_name as level3 from people as p left join department as d1 on d1.dep_id = p.dep_id left join department as d2 on d2.dep_id = d1.parent_id left join department as d3 on d3.dep_id = d2.parent_id where d1.dep_name='B' or d2.dep_name='B' or d3.dep_name='B';
Find the total number of people whose secondary department is B
select count(*) as total from people as p left join department as d1 on d1.dep_id = p.dep_id left join department as d2 on d2.dep_id = d1.parent_id left join department as d3 on d3.dep_id = d2.parent_id where d1.dep_name='B' or d2.dep_name='B' or d3.dep_name='B';
| department_id | departname | ||
|---|---|---|---|
| 4 | D | 2 | |
| 5 | E | 3 | |
| 6 | F | 4 | |
| 2 | B | 5 | |
| 2 | B |
Second question: Simple statistics
CREATE TABLE STUDENT(
ID INT UNSIGNED AUTO_INCREMENT,
SCORE1 INT NOT NULL,
SCORE2 INT NOT NULL,
SCORE3 INT NOT NULL,
SCORE4 INT NOT NULL,
PRIMARY KEY(ID)
)CHARSET=utf8;Copy after login
CREATE TABLE STUDENT(
ID INT UNSIGNED AUTO_INCREMENT,
SCORE1 INT NOT NULL,
SCORE2 INT NOT NULL,
SCORE3 INT NOT NULL,
SCORE4 INT NOT NULL,
PRIMARY KEY(ID)
)CHARSET=utf8;INSERT INTO STUDENT VALUES(1,100,98,10,4),(2,100,9,10,4),(3,70,0,180,40),(4,10,98,1,4),(5,30,7,10,4),(6,8,88,1,43);
SELECT id, score1,score2,score3,score4, score1+score2+score3+score4 as total FROM STUDENTwhere score1+score2+score3+score4 > 200 order by score1+score2+score3+score4 desc;
Here is a question that cannot be directly Knowledge points for sorting by alias,
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