Crud of MySQL

Catalog

Preface

One 、 The basic query

Two 、 Filter query

3、 ... and 、 Sort

Four 、 Case study

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Preface

Last time I shared MySQL Account management , What we want to share this time is CRUD, Anyway CRUD There must be some .

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One 、 The basic query

#1. A single field in a query table 
SELECT department_id FROM t_mysql_employees
	#2. Multiple fields in the query table 
	SELECT department_id,employee_id FROM t_mysql_employees

	#3. Query all fields in the table 
	SELECT * FROM t_mysql_employees

	#4. Query constant value 
	SELECT 100;
	SELECT 'john';
	#5. Query expression 
	SELECT 100%98;
	#6. Query function 
	SELECT VERSION();
	#7. names 
	SELECT department_id  Department number  FROM t_mysql_employees

	#8. duplicate removal 
	SELECT DISTINCT department_id FROM t_mysql_employees

	​# Case study ： Query all department numbers involved in the employee table 
		SELECT DISTINCT department_id FROM t_mysql_employees

	#9.+ The role of No 
	SELECT 'join'+1
	SELECT '100'+1

Running results ：  

  #1. A single field in a query table

    #2. Multiple fields in the query table

    #3. Query all fields in the table

    #4. Query constant value
    SELECT 100;

    SELECT 'john';

    #5. Query expression
    SELECT 100%98;

    #6. Query function
    SELECT VERSION();

    #7. names

    #8. duplicate removal

    #9.+ The role of No

 

 

 

Two 、 Filter query

# One 、 Filter by conditional expression 
    # Case study 1： Query salary >12000 Employee information 
SELECT * FROM t_mysql_employees WHERE salary > 12000

    # Case study 2： Query department number is not equal to 90 The employee name and department number of the 
SELECT first_name,department_id FROM t_mysql_employees WHERE NOT(department_id=90)
    ​
    # Two 、 Filter by logical expression 
    # Case study 1： Query salary z stay 10000 To 20000 Between the employee's name 、 Salary and bonus 
SELECT first_name,salary,commission_pct FROM t_mysql_employees WHERE salary BETWEEN 10000 AND 20000

    # Case study 2： The inquiry department number is not in 90 To 110 Between , Or pay more than 15000 Employee information 
SELECT * FROM t_mysql_employees WHERE NOT(department_id BETWEEN 90 AND 110) or salary > 12000


    # 3、 ... and 、 Fuzzy query 
    #1.like
    # Case study 1： Query employee name contains characters a Employee information 
SELECT * FROM t_mysql_employees WHERE first_name LIKE '%a%'

    # Case study 2： The third character in the query employee name is e, The fifth character is a The name and salary of the employee 
SELECT * FROM t_mysql_employees WHERE first_name LIKE '__e_a%'

    # Case study 3： The second character in the employee name is _ Employees 
SELECT * FROM t_mysql_employees WHERE last_name LIKE 

    #2.between and
    # Case study 1： Query the employee number in 100 To 120 Employee information between 
SELECT * FROM t_mysql_employees WHERE employee_id BETWEEN 100 AND 120

    #3.in
    # Case study ： The job number of the employee is  IT_PROG、AD_VP、AD_PRES An employee's name and job number in the 
SELECT first_name,job_id FROM t_mysql_employees WHERE job_id IN ('IT_PROG','AD_VP','AD_PRES') 

    #4、is null
    # Case study 1： Query the employee name and bonus rate without bonus 
SELECT first_name,commission_pct FROM t_mysql_employees WHERE commission_pct is null
    # Case study 2： Check the name and bonus rate of employees with bonus 
SELECT first_name,commission_pct FROM t_mysql_employees WHERE not commission_pct is null
    # Safety is equal to  <=>
    ​# Case study 1： Query the employee name and bonus rate without bonus 
SELECT first_name,commission_pct FROM t_mysql_employees WHERE commission_pct <=> null

    # Case study 2： Query salary as 12000 Employee information 
SELECT * FROM t_mysql_employees WHERE salary <=>12000 

    #is null pk <=>
    IS NULL: Can only judge NULL value , High readability , It is recommended to use 
    <=>   : You can judge NULL value , You can also judge ordinary values , Less readable 

One 、 Filter by conditional expression
    # Case study 1： Query salary >12000 Employee information

    # Case study 2： Query department number is not equal to 90 The employee name and department number of the
    ​
    # Two 、 Filter by logical expression
    # Case study 1： Query salary z stay 10000 To 20000 Between the employee's name 、 Salary and bonus

    # Case study 2： The inquiry department number is not in 90 To 110 Between , Or pay more than 15000 Employee information

 

 

 

 

    # 3、 ... and 、 Fuzzy query
    #1.like
    # Case study 1： Query employee name contains characters a Employee information

    # Case study 2： The third character in the query employee name is e, The fifth character is a The name and salary of the employee

 

    #2.between and
    # Case study 1： Query the employee number in 100 To 120 Employee information between

    #3.in
    # Case study ： The job number of the employee is IT_PROG、AD_VP、AD_PRES An employee's name and job number in the

 

 

    #4、is null
    # Case study 1： Query the employee name and bonus rate without bonus

    # Case study 2： Check the name and bonus rate of employees with bonus

 

 

    # Safety is equal to <=>
    ​# Case study 1： Query the employee name and bonus rate without bonus

    # Case study 2： Query salary as 12000 Employee information
    ​
    #is null pk <=>
    IS NULL: Can only judge NULL value , High readability , It is recommended to use
    <=>   : You can judge NULL value , You can also judge ordinary values , Less readable

 
 

3、 ... and 、 Sort

order by  Clause 
    ​#1、 Sort by a single field 
    # Case study ： Sort by employee table salary 
SELECT * FROM t_mysql_employees ORDER BY salary    ​

    #2、 Add filter criteria and sort 
    ​# Case study ： Check department number >=90 Employee information , And in descending order of employee number 
SELECT * FROM t_mysql_employees WHERE department_id>=90 ORDER BY employee_id DESC    ​

    #3、 Sort by expression 
    # Case study ： Query employee information   In descending order of annual salary 
SELECT em.salary*12*(1+IFNULL(commission_pct,0)),em.* FROM t_mysql_employees em ORDER BY em.salary*12*(1+IFNULL(commission_pct,0)) DESC
    ​​
    #4、 Sort by alias 
    # Case study ： Query employee information   In ascending order of annual salary 
SELECT em.salary*12*(1+IFNULL(commission_pct,0)),em.* FROM t_mysql_employees em ORDER BY em.salary*12*(1+IFNULL(commission_pct,0)) ASC   

    #5、 Sort by function 
    # Case study ： Query employee name , And in descending order of the length of the names 
SELECT first_name FROM t_mysql_employees ORDER BY LENGTH(first_name) DESC
    ​​
    #6、 Sort by multiple fields 
    ​# Case study ： Query employee information , It is required that the salary be in descending order first , Press again employee_id Ascending 
SELECT * FROM t_mysql_employees ORDER BY salary DESC,employee_id ASC


#1. Check the employee's name, department number and annual salary , In descending order of annual salary   Ascending by name 
SELECT em.first_name,em.department_id,em.salary*12*(1+IFNULL(commission_pct,0)) FROM t_mysql_employees em ORDER BY em.salary*12(1+IFNULL(commission_pct,0)) DESC,first_name asc

	#2. The choice of salary is not in 8000 To 17000 The name and salary of the employee , In descending order of wages 
	SELECT first_name,salary FROM t_mysql_employees WHERE NOT(salary BETWEEN 8000 AND 17000) ORDER BY salary desc
	
	#3. Query mailbox contains e Employee information , And first by the number of bytes in the mailbox , And then in ascending order by department number 
	SELECT * FROM t_mysql_employees WHERE email LIKE '%e%' ORDER BY LENGTH(email)DESC,department_id ASC

order by Clause
    ​#1、 Sort by a single field
    # Case study ： Sort by employee table salary
    ​
    #2、 Add filter criteria and sort
    ​# Case study ： Check department number >=90 Employee information , And in descending order of employee number
    ​
    #3、 Sort by expression
    # Case study ： Query employee information In descending order of annual salary
    ​​
    #4、 Sort by alias
    # Case study ： Query employee information In ascending order of annual salary
    ​​
    #5、 Sort by function
    # Case study ： Query employee name , And in descending order of the length of the names
    ​​
    #6、 Sort by multiple fields
    ​# Case study ： Query employee information , It is required that the salary be in descending order first , Press again employee_id Ascending

 

 

 

  

#1. Check the employee's name, department number and annual salary , In descending order of annual salary Ascending by name

    #2. The choice of salary is not in 8000 To 17000 The name and salary of the employee , In descending order of wages

 

    #3. Query mailbox contains e Employee information , And first by the number of bytes in the mailbox , And then in ascending order by department number

 

 

Four 、 Case study

·insert into t_student values('01' , ' Zhao Lei ' , '1990-01-01' , ' male ');
insert into t_student values('02' , ' Qian Dian ' , '1990-12-21' , ' male ');
insert into t_student values('03' , ' Sun Feng ' , '1990-12-20' , ' male ');
insert into t_student values('04' , ' Li Yun ' , '1990-12-06' , ' male ');
insert into t_student values('05' , ' Zhou Mei ' , '1991-12-01' , ' Woman ');
insert into t_student values('06' , ' Wu Lan ' , '1992-01-01' , ' Woman ');
insert into t_student values('07' , ' Zheng Zhu ' , '1989-01-01' , ' Woman ');
insert into t_student values('09' , ' Zhang San ' , '2017-12-20' , ' Woman ');
insert into t_student values('10' , ' Li Si ' , '2017-12-25' , ' Woman ');
insert into t_student values('11' , ' Li Si ' , '2012-06-06' , ' Woman ');
insert into t_student values('12' , ' Zhao Liu ' , '2013-06-13' , ' Woman ');
insert into t_student values('13' , ' Sun Qi ' , '2014-06-01' , ' Woman ');

--  Teachers list 
insert into t_teacher values('01' , ' Zhang San ');
insert into t_teacher values('02' , ' Li Si ');
insert into t_teacher values('03' , ' Wang Wu ');

--  The curriculum 
insert into t_course values('01' , ' Chinese language and literature ' , '02');
insert into t_course values('02' , ' mathematics ' , '01');
insert into t_course values('03' , ' English ' , '03');

--  League tables 
insert into t_score values('01' , '01' , 80);
insert into t_score values('01' , '02' , 90);
insert into t_score values('01' , '03' , 99);
insert into t_score values('02' , '01' , 70);
insert into t_score values('02' , '02' , 60);
insert into t_score values('02' , '03' , 80);
insert into t_score values('03' , '01' , 80);
insert into t_score values('03' , '02' , 80);
insert into t_score values('03' , '03' , 80);
insert into t_score values('04' , '01' , 50);
insert into t_score values('04' , '02' , 30);
insert into t_score values('04' , '03' , 20);
insert into t_score values('05' , '01' , 76);
insert into t_score values('05' , '02' , 87);
insert into t_score values('06' , '01' , 31);
insert into t_score values('06' , '03' , 34);
insert into t_score values('07' , '02' , 89);
insert into t_score values('07' , '03' , 98);









#01） Inquire about " 01 " Course than " 02" Information and course marks for students with high course grades 
SELECT  st.*,sc.score AS '01',sc2.score AS '02' FROM t_student st
LEFT JOIN t_score sc ON sc.sid=st.sid AND sc.cid='01'
LEFT JOIN t_score sc2 ON sc2.sid=st.sid AND sc2.cid='02'
WHERE sc.score>sc2.score

#02） Queries exist at the same time " 01 " Courses and " 02 " Course situation 
SELECT st.* FROM t_student st INNER JOIN t_score sc ON sc.sid=st.sid  AND sc.cid='01'
WHERE st.sid IN (
SELECT st2.sid FROM t_student st2 INNER JOIN t_score sc2 ON sc2.sid=st2.sid AND sc2.cid='02'
)

#03） Query exists " 01 " The course may not exist " 02 " Course situation ( If it does not exist, it will be displayed as  null )
SELECT st.* FROM t_student st INNER JOIN t_score sc ON sc.sid=st.sid  AND sc.cid='01'
WHERE st.sid NOT IN (
SELECT st2.sid FROM t_student st2 INNER JOIN t_score sc2 ON sc2.sid=st2.sid AND sc2.cid='02'
)


#04） Query does not exist " 01 " Course but there is " 02 " Course situation 
SELECT * FROM t_student st INNER JOIN t_score sc ON sc.sid=st.sid AND sc.cid='02'
WHERE st.sid NOT IN (
SELECT st2.sid FROM t_student st2 INNER JOIN t_score sc2 ON sc2.sid=st2.sid AND sc2.cid='01'
)


#05） Query average score is greater than or equal to  60  Student number, student name and average score of each student 
SELECT st.sid,st.sname,ROUND(AVG(sc.score),2)  average 
FROM  t_student st 
LEFT JOIN t_score sc ON st.sid=sc.sid
GROUP BY st.sid HAVING AVG(sc.score)>=60;


#06） The query in t_score There is student information about the scores in the table 
SELECT st.* FROM t_student st WHERE sid IN(
SELECT sc.sid FROM t_score sc
)

#07） Query student numbers of all students 、 The student's name 、 The total number of selected courses 、 Total grade of all courses ( No results are shown as  null )
SELECT st.sid,st.sname,COUNT(sc.cid),ROUND(SUM(sc.score),2) AS ' Total score ' FROM  t_student st 
LEFT JOIN t_score sc ON sc.sid=st.sid GROUP BY st.sid



#08） Inquire about 「 Li 」 The number of teachers surnamed 
SELECT COUNT(*) FROM t_teacher WHERE tname LIKE ' Li %'


#09） I have learned to query 「 Zhang San 」 Information of students taught by teachers 
SELECT st.* FROM t_student st WHERE sid IN(
SELECT sid FROM t_score WHERE cid = (
SELECT cid FROM t_course WHERE tid = (
SELECT tid FROM t_teacher WHERE tname = ' Zhang San '
)
)
)

#10） Check the information of students who have not learned all the courses 
SELECT st.*FROM  t_student st WHERE sid NOT IN(
SELECT sid FROM t_score GROUP BY sid HAVING COUNT(cid)>=(SELECT COUNT(cid) FROM t_course)
)

#11） I didn't learn how to query " Zhang San " The name of the student in any course taught by the teacher 
SELECT st.sname FROM t_student st WHERE st.sid NOT IN (
SELECT sc.sid FROM t_score sc 
INNER JOIN t_course c ON c.cid=sc.cid
INNER JOIN t_teacher t ON t.tid=c.tid AND t.tname=" Zhang San "
)

#12） Check the student number of two or more failed courses , Name and average score 
SELECT st.sid,st.sname,AVG(sc.score) FROM t_student st
LEFT JOIN t_score sc ON sc.sid=st.sid WHERE sc.sid IN(
SELECT sc.sid FROM t_score sc WHERE sc.score<60 OR sc.score IS NULL GROUP BY sc.sid HAVING COUNT(1)>=2
)
GROUP BY st.sid


#13） retrieval " 01 " The course score is less than  60, Student information in descending order of scores 
SELECT st.*,sc.score FROM t_student st LEFT JOIN  t_score sc ON sc.sid=st.sid WHERE sc.cid='01' AND sc.score<60 ORDER BY sc.score DESC ;

#14） Show the grades of all courses and the average grades of all students from high to low 
SELECT st.sid,st.sname,
(IFNULL((sc4.score),0)) " average ",
(IFNULL((sc.score),0)) " Chinese language and literature ",
(IFNULL((sc2.score),0)) " mathematics ",
(IFNULL((sc3.score),0))" English " 
FROM t_student st
LEFT JOIN t_score sc  ON sc.sid=st.sid  AND sc.cid="01"
LEFT JOIN t_score sc2 ON sc2.sid=st.sid AND sc2.cid="02"
LEFT JOIN t_score sc3 ON sc3.sid=st.sid AND sc3.cid="03"
LEFT JOIN t_score sc4 ON sc4.sid=st.sid
GROUP BY st.sid 
ORDER BY AVG(sc4.score) DESC



#15） Check the highest scores of all subjects 、 Minimum and average points ：
# Show as follows ： Course  ID, Course  name, The highest , Lowest score , average , pass rate , Medium rate , Excellent rate , The pass rate of excellence is >=60, The average is ：70-80, Good is ：80-90, Excellence is ：>=90
# Require output of course number and number of electives , The query results are arranged in descending order of the number of people , If the number of people is the same , In ascending order of course number 

SELECT c.cid,c.cname,MAX(sc.score) " The highest ",MIN(sc.score) " Lowest score ",AVG(sc.score) " average " 
,((SELECT COUNT(sid) FROM t_score WHERE score>=60 AND cid=c.cid )/(SELECT COUNT(sid) FROM t_score WHERE cid=c.cid)) " pass rate "
,((SELECT COUNT(sid) FROM t_score WHERE score>=70 AND score<80 AND cid=c.cid )/(SELECT COUNT(sid) FROM t_score WHERE cid=c.cid)) " Medium rate "
,((SELECT COUNT(sid) FROM t_score WHERE score>=80 AND score<90 AND cid=c.cid )/(SELECT COUNT(sid) FROM t_score WHERE cid=c.cid)) " Excellent rate "
,((SELECT COUNT(sid) FROM t_score WHERE score>=90 AND cid=c.cid )/(SELECT COUNT(sid) FROM t_score WHERE cid=c.cid)) " Excellence rate "
FROM t_course c
LEFT JOIN t_score sc ON sc.cid=c.cid 
GROUP BY c.cid

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