[database] SQL Server quickly creates tables to simulate departments, courses, teachers, students and scores

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In this article , It is mainly about creating tables and how to quickly simulate table data , Convenient and fast for query testing and verification
Compared with addition, deletion and modification , Queries for any system , Are the main functions , And some operations can only be carried out after there is a certain amount of data , such as ： Statistics of average scores of students, etc

• List of main knowledge points

Number	Language or plug-in	Knowledge point	explain
1	sql server	create table	key word , Create table
2	sql server	indentity	Self growth function
3	sql server	primary key	Set the primary key id
4	sql server	len()	Statistical length
5	sql server	rand()	Random function ,0~1 Between the value of the
6	sql server	substring()	Intercepting string ,substring('',1,1)
7	sql server	convert()	Data type conversion ,convert(int,'')
8	sql server	left join	Make connections , Mainly the left table , There will be 3 * 5=15 The situation of this record , It needs to be used properly on Conditions
9	sql server	avg()	averaging

【 Query exercise requirements 】

1） List presence 86 Name and course number of students with above grades

2） List “C Language ” Students whose grades are lower than average

【 Table design and analysis 】

1） Pass the above requirements , The following can be designed simply 5 A watch , Omit the class table for the time being

2） Department table

Department number 、 Department name

• Create table

create table system_table
(
    system_id int identity(1,1) primary key,
    system_name nvarchar(50)
)

• Analog data

-- Suppose the school has the following department data ： department of business 、 Management Department 、 Department of Computer Science 、 Architecture Department 、 Art Department ,5 Personal records 
insert into system_table(system_name)
values(' department of business '),(' Management Department '),(' Department of Computer Science '),(' Architecture Department '),(' Art Department ')

3） The curriculum Department number 、 Course number 、 Course name

create table course_table
(
    course_id int identity(1,1) primary key,
    course_name nvarchar(50),
    system_id int
)

• Analog data

-- At this time, the data is different , There will be associated tables , Different departments have different courses 
-- Here we only simulate the computer system 3 Course ,C Language 、 Software Engineering 、 database 
insert into course_table(course_name,system_id)
values('C Language ',3),(' Software Engineering ',3),(' database ',3)

3） Teachers list

Department number 、 Course number 、 Teacher number 、 Teacher's name

create table teacher_table
(
    teacher_id int identity(1,1) primary key,
    teacher_name nvarchar(50),
    course_id int,
    system_id int
)

• Analog data

-- It is also related , Simulate here 3 Bar record , Zhang San 、 Li Si 、 Wang Wu 
insert into teacher_table(teacher_name,course_id,system_id)
values(' Zhang San ',1,3),(' Li Si ',2,3),(' Wang Wu ',3,3)

4） Student list

Department number 、 Course number 、 Teacher number 、 Student number 、 The student's name 、 Gender

create table student_table
(
    student_id int identity(1,1) primary key,
    student_name nvarchar(50),
    sex int, --1= schoolboy ,0= girl student 
    teacher_id int,
    course_id int,
    system_id int
)

• The data simulation

declare @myName nvarchar(50)
declare @surname_arr nvarchar(1000)
declare @name_arr nvarchar(1000)
set @surname_arr=' Wang Li Zhang Liu Chen Yang Huang Zhao Wu Zhou Xu sun Ma Zhu Hu Guo '
set @name_arr=' Love dream box Yu Yu Si Pei poem purple string sharp moon summer phantom shadow dark according to dancing Ling drop Yu Yingying '

declare @sn_length int
declare @rand_index int

-- Analog data ： Department of Computer Science 、C Language courses 、 Mr. Zhang San 30 Famous student 
declare @student_count int
set @student_count=30
declare @index int 
set @index=0
declare @sex int
set @sex=1

while @index<@student_count begin
	set @myName='' -- Initial value of Fu , Otherwise it won't work + plus 
	
	-- Get the last name value 
	if len(@surname_arr)>0 begin
		set @sn_length=len(@surname_arr)
		set @rand_index=convert(int,(rand()*@sn_length))-1
		set @myName+=substring(@surname_arr,@rand_index,1)
	end

	-- Get the name 
	if len(@name_arr)>0 begin
		set @sn_length=len(@name_arr)
		set @rand_index=convert(int,(rand()*@sn_length))-1
		set @myName+=substring(@name_arr,@rand_index,2)
	end
	
	-- Gender 
	set @sex=convert(int,(rand()*2))
	
	insert into student_table(student_name,sex,teacher_id,course_id,system_id)
	values(@myName,@sex,1,1,3)
	
	set @index+=1
end

• Inquire about

select a.*,b.teacher_name,c.course_name,d.system_name
from student_table a
left join teacher_table b on b.teacher_id=a.teacher_id
left join course_table c on c.course_id=a.course_id
left join system_table d on d.system_id=a.system_id

5） Student transcript

Student number 、 Department number 、 Course number 、 Teacher number 、 fraction

create table student_score_table
(
    score_id int identity(1,1) primary key,
    score int,
    student_id int,
    teacher_id int,
    course_id int,
    system_id int
)

• Analog data

declare @index int
set @index=1
declare @score int
set @score=0

while @index<=30 begin
	
	set @score=convert(int,(rand()*50))+50
	
	insert into student_score_table(score,student_id,teacher_id,course_id,system_id)
	values(@score,@index,1,1,3)

	set @index+=1
end

• Inquire about

select a.*,a2.student_name,b.teacher_name,c.course_name,d.system_name
from student_score_table a
inner join student_table a2 on a2.student_id=a.student_id
left join teacher_table b on b.teacher_id=a.teacher_id
left join course_table c on c.course_id=a.course_id
left join system_table d on d.system_id=a.system_id

【 Formal inquiry of the requirements mentioned above 】

1） List presence 86 Name and course number of students with above grade

select a.score,c.student_name,a.course_id,b.course_name
from student_score_table a
left join course_table b on b.course_id=a.course_id
left join student_table c on c.student_id=a.student_id
where a.score>86

2） List “C Language ” Students whose grades are lower than average

It's used here , The average score of the inline sub query is

select a.score,b.avg_score,a.student_id,c.student_name
from student_score_table a
left join(
	--C Average score of language courses 
	select course_id,avg(score) as avg_score
	from student_score_table 
	where course_id=1
	group by course_id
) as b on b.course_id=a.course_id
left join student_table c on c.student_id=a.student_id
where a.score<b.avg_score