MySQL query related knowledge (advanced level 7: sub query

classification ：

Click the location where the subquery appears ：

select Back ：

Only scalar subqueries are supported

from Back ：

Support table subquery

where or having Back *

Scalar subquery （ A single ）√

Column query （ Multiple lines ）√

Line sub query

exists Back （ Correlation subquery ）

Table sub query

Depending on the number of rows and columns in the result set ：

Scalar subquery （ The result set has only one row and one column ）

Column query （ The result set has only one row and more columns ）

Line sub query （ The result set has one row and more columns ）

Table sub query （ The result set is usually multi row and multi column ）

[](() One 、where or having Back

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1. Scalar subquery （ Single line sub query ）

2. Column query （ Multi line sub query ）

3. Line sub query （ Multiple rows and columns ）

characteristic ：

1. Subqueries in parentheses

2. Subqueries are usually placed on the right side of the condition

3. Scalar subquery , Usually with a single line operator （>,<,= wait ） Use

Column query , Generally, this multi line operator is used （in,any/some,all） Use

4. The execution of the subquery takes precedence over the execution of the main query , The conditions of the main query use the results of the sub query

*/

[](()1. Scalar subquery

# Case study 1、 Whose pay ratio is Abel high ？

#① Inquire about Abel The salary of

SELECT salary

FROM employees

WHERE last_name=‘Abel’;

#② Search for employee information , Satisfy salary>① result

SELECT *

FROM employees

WHERE salary>(

SELECT salary

FROM employees

WHERE last_name=‘Abel’

);

# Case study 2： return job_id And 141 Same as employee No ,salary Than 143 Employees with more employees full name ,job_id And wages

SELECT last_name,job_id,salary

FROM employees

WHERE job_id=(

SELECT job_id

FROM employees

WHERE employee_id=141

)AND salary>(

SELECT salary

FROM employees

WHERE employee_id=143

);

# Case study 3： Return the name of the employee with the lowest salary in the company ,job_id and salary

SELECT last_name,job_id,salary

FROM employees

WHERE salary=(

SELECT MIN(salary)

FROM employees

);

# Case study 4、 The minimum wage is greater than 50 The Department of minimum wage id And its minimum wage

SELECTdepartment_id,MIN(salary)

FROM employees

GROUP BY department_id

HAVING MIN(salary)>(

SELECT MIN(salary)

FROM employees

WHERE department_id=50

);

[](()2. Column query （ Multi line sub query ）

# Case study 1、 return location_id yes 1400 or 1700 The names of all employees in the Department

SELECT last_na **《 A big factory Java Analysis of interview questions + Back end development learning notes + The latest architecture explanation video + Practical project source code handout 》 Free open source Prestige search official account 【 Advanced programming 】** me

FROM employees

WHERE department_id IN(

SELECT DISTINCT department_id

FROM departments

WHERE location_id IN (1400,1700)

);

# Case study 2、 Return to other types of work job_id by ‘IT＿PROG’ Employees of any type of work with low wages ： Job number , full name ,job_id as well as salary

SELECT employee_id,last_name,job_id,salary

FROM employees

WHERE salary < ANY(

SELECT DISTINCT salary

FROM employees

WHERE job_id=‘IT_PROG’

)

AND job_id<>‘IT_PROG’;

# Case study 3、 Return to other types of work job_id by ‘IT＿PROG’ Of all low paid employees in the type of work ： Job number , full name ,job_id as well as salary

SELECT employee_id,last_name,job_id,salary

FROM employees

WHERE salary < ALL(

SELECT DISTINCT salary

FROM employees

WHERE job_id=‘IT_PROG’

)

AND job_id<>‘IT_PROG’;

perhaps

SELECT employee_id,last_name,job_id,salary

FROM employees

WHERE salary < (

SELECT MIN(salary)

FROM employees

WHERE job_id=‘IT_PROG’

)

AND job_id<>‘IT_PROG’;

[](()3. Line sub query

（ The result set has one row and multiple columns or multiple rows and multiple columns ）( When using this method, all information can be represented by a judgment symbol , This method can also be obtained by other subqueries )

# Case study 、 Query the information of the employee with the lowest employee number and the highest salary

SELECT *

FROM employees

WHERE (employee_id,salary)=(

SELECT MIN(employee_id),MAX(salary)

FROM employees

);

[](() Two 、 Put it in SELECT Behind ( Only scalar subqueries are supported （ That is, one row, one column ）)

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# Case study 、 Query the number of employees in each department

SELECT d.*,(

SELECT COUNT(*)

FROM employees e

WHERE e.department_id=d.department_id

) Number

FROM departments d;

# Case study 、 Check the employee number =102 The name of your department

SELECT department_name

FROM departments d,employees e

WHERE d.department_id=e.department_id

AND e.employee_id=102;

[](() 3、 ... and 、 Put it in FROM Behind ( This means that the result of the query is used as a table （ The alias is required ）)

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# Case study 、 Query the average salary grade of each department

SELECT biao.*,j.grade_level

FROM (

SELECT AVG(salary) ag,department_id

FROM employees

GROUP BY department_id

)biao

INNER JOIN job_grades j

ON biao.ag BETWEEN j.lowest_sal AND highest_sal;

[](() Four 、EXISTS Back （ Correlation subquery ）

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