MySQL—— Subquery usage

1、 Overview of subquery

Introduction to subquery

Subquery refers to a query that is nested within another query statement by one query statement , Internal query is the condition of external query , This feature comes from MySQL4.1 Start introducing .

SQL The use of neutron queries has greatly enhanced SELECT The ability to query , Because many times queries need to get data from the result set , Or you need to calculate a data result from the same table first , Then with this data result （ It could be some scalar , It could also be a collection ） Compare .

Subquery basically uses

grammar ：

SELECT --  Main query 
    select_list
FROM
    table
WHERE
    expr operator > (SELECT --  Subquery 
            select_list
        FROM
            table);

Subquery ( Internal query ) The execution is completed before the main query .

The results of the sub query are by the main query ( External query ) Use .

matters needing attention ：

• Subqueries should be enclosed in parentheses

• Place the subquery to the right of the comparison criteria

• Single line operators correspond to single line subqueries , Multiline operators correspond to multiline subqueries

Classification of subqueries

Return one or more records according to the results of sub query , Sub query is divided into single line sub query and multi line sub query .

Whether the sub query is executed more than once , Divide the subquery into related ( Or associated ) Subquery and unrelated ( Or unrelated ) Subquery .

2、 Use of subquery

2.1、 Single line sub query

Single line comparison operator

= be equal to ,> Greater than ,>= Greater than or equal to ,< Less than ,<= Less than or equal to ,<> It's not equal to

Using examples

--  Find employees whose salary is greater than the average salary of all employees 
SELECT employee_id, first_name, last_name, salary
FROM employees
WHERE salary > (
    SELECT AVG(salary) FROM employees);

--  Query salary greater than 149 Information of the employee whose salary is No 
select employee_id,last_name,salary
from employees
where salary > (
    select salary from employees where employee_id = 149)
   
--  return job_id And 141 Same as employee No ,salary Than 143 The name of the employee with a large number of employees ,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
    );

--  Return to the lowest paid employees of the company last_name,job_id and salary
select last_name,job_id,salary
from employees
where salary = (
    select min(salary) from employees
    );

--  Query and 147 Of employee No manager_id and department_id Of the same other employees employee_id,manager_id,department_id
select employee_id,manager_id,department_id
from employees
where manager_id = (
    select manager_id from employees where employee_id = 147
    )
and department_id = (
    select department_id from employees where employee_id = 147
    )
and employee_id <> 147;    
    
--  The minimum wage is greater than 50 The Department of minimum wage id And its minimum wage 
select department_id,min(salary) min_salary
from employees
where department_id is not null
group by department_id
having min_salary > (
    select min(salary) from employees where department_id = 50
    );
    
    

2.2、 Multi line sub query

Multi row subquery comparison operator

Using examples

--  Find the location that belongs to ID by 1700 All employees 
SELECT
    employee_id, first_name, last_name
FROM
    employees
WHERE
    department_id IN (SELECT
            department_id
        FROM
            departments
        WHERE
            location_id = 1700)
ORDER BY first_name , last_name;

--  Return to other job id Middle ratio job id by 'IT_PROG’ Employee number and name of any low paid employee in the Department 、job id as well as salary
select last_name,job_id,salary
from employees
where salary < any (
    select salary from employees where job_id = 'IT_PROG'
)
and job_id <> 'IT_PROG';

--  Return to other job id Middle ratio job id by 'IT_PROG’ Employee number and name of all low paid employees in the Department 、job id as well as salary
select last_name,job_id,salary
from employees
where salary < all (
    select salary from employees where job_id = 'IT_PROG'
)
and job_id <> 'IT_PROG';

--  Query the Department with the lowest average wage ID
--  Mode one ：
select department_id,avg(salary) from employees group by department_id order by avg(salary) limit 1;
--  Mode two ：
select department_id
from employees
group by department_id
having avg(salary) = (
    select min(avg_sal)
    from (
     select avg(salary) avg_sal from employees group by department_id
         ) t_dept_avg_sal
    )
--  Mode three ：
select department_id
from employees
group by department_id
having avg(salary) <= all (
    select avg(salary) from employees group by department_id
);

--  Query the employee information of the leader in the employee table 
select employee_id,last_name,manager_id
from employees
where employee_id in (
    select manager_id from employees
    );
    
--  Query the information of employees who are not leaders in the employee table 
select employee_id,last_name,manager_id
from employees
where employee_id not in (
    select manager_id from employees where manager_id is not null
    );

2.3、 Correlation subquery

If the execution of a subquery depends on an external query , Usually, it is because the tables in the subquery use external tables , Conditional correlation is carried out , So every time you execute an external query , All subqueries have to be recalculated , Such a subquery is called an associated subquery .

Related sub queries are executed row by row , Each row of the main query executes a subquery .

Using examples

--  Query the salary of employees whose salary is greater than the average salary of the Department last_name, salary And its department_id
--  Mode one 
select last_name,salary,department_id
from employees e1
where salary > (
    select avg(salary) from employees e2 where e1.department_id = e2.department_id
    );
--  Mode two 
select last_name,salary,e.department_id
from employees e ,(select department_id,avg(salary) avg_sal from employees group by department_id) t_dept_avg_sal
where e.department_id = t_dept_avg_sal.department_id
and e.salary > t_dept_avg_sal.avg_sal;

--  Check the employee's id,salary, according to department_name Sort 
select employee_id,salary
from employees e
order by (
    select department_name from departments d where e.department_id = d.department_id
             );
                     

EXISTS And NOT EXISTS

EXISTS Operator is used to specify whether there are qualified rows in the subquery ,

If the subquery contains any rows , be EXISTS Operator return true. Otherwise it returns false.

EXISTS Operator terminates query processing immediately after a row is found , therefore , You can use EXISTS Operator to improve query performance

NOT EXISTS If there is no certain condition , Then return to true, Otherwise return to true.

Using examples ：

--  Find manager's id, full name , Work and department id
select employee_id,last_name,job_id,department_id
from employees e1
where exists(
    select * from employees e2 where e1.employee_id = e2.manager_id
          )
          
--  Inquire about departments In the table , There is no in employees Of the departments in the table department_id and department_name
--  Mode one 
select department_id,department_name
from departments d
where not exists(
    select * from employees e where d.department_id = e.department_id
    );
--  Mode two 
select d.department_id,department_name
from departments d left join employees e on d.department_id = e.department_id
where e.department_id is null;          
          

Subquery exercises

#1. Query and Zlotkey Name and salary of employees in the same department 
select last_name,salary
from employees
where department_id in (
    select department_id from employees where last_name = 'Zlotkey'
    )
and last_name <> 'Zlotkey';

#2. Query the employee number of the employee whose salary is higher than the average salary of the company , Name and salary .
select employee_id,last_name,salary
from employees
where salary > (
    select avg(salary) from employees
    );

#3. Select salary greater than all JOB_ID = 'SA_MAN' The salary of the employee last_name,job_id, salary
select last_name,job_id,salary
from employees
where salary > (
    select max(salary) from employees where job_id = 'SA_MAN'
    );

#4. Queries and names contain letters u Employee number and name of employees in the same department 
select employee_id,last_name
from employees
where department_id in (
    select department_id from employees where last_name like '%u%'
    );

#5. Check the... In the Department location_id by 1700 The employee number of the employee working in the Department 
select employee_id
from employees e
where department_id in (
    select department_id from departments where location_id = 1700
    );

#6. The query manager is King The name and salary of the employee 
select last_name,salary
from employees
where  manager_id in (
    select employee_id from employees where last_name = 'K_ing'
    );


#7. Query the lowest paid employee information : last_name,salary
select last_name,salary
from employees
order by salary
limit 1;

select last_name,salary
from employees
where salary = (
    select min(salary) from employees
    );

#8. Search the Department with the lowest average wage 
--  Mode one ：
select * from departments
where department_id = (
    select department_id
    from employees
    group by department_id
    order by avg(salary)
    limit 1
    );
--  Mode two ：
select * from departments
where department_id = (
    select department_id
    from employees
    group by department_id
    having avg(salary) = (
       select min(avg_sal) from (select avg(salary) avg_sal from employees group by department_id) t_sal
    )
);
--  Mode three ：
select * from departments
where department_id = (
    select department_id
    from employees
    group by department_id
    having avg(salary) <= all (select avg(salary) from employees group by department_id)
    );

#9. Query the information of the Department with the lowest average wage and the average wage of the Department , Subqueries can be written in the query field 
select d.* ,(select avg(salary) from employees where department_id = d.department_id) avg_sal
from departments d
where department_id = (
    select department_id
    from employees
    group by department_id
    having avg(salary) <= all (select avg(salary) from employees group by department_id)
    );

#10. Find the highest average wage job  Information 
--  Mode one ：
select * from jobs
where job_id = (
    select job_id
    from employees
    group by job_id
    having avg(salary) >= all (
        select avg(salary) from employees group by job_id )
    );
--  Mode two ：
select * from jobs
where job_id = (
    select job_id
    from employees
    group by job_id
    order by avg(salary) desc
    limit 1
    );

#11. Query the departments whose average salary is higher than the average salary of the company ?
select distinct department_id
from employees
where department_id is not null
group by department_id
having avg(salary) > (
    select avg(salary) from employees
    );

#12. Find out all the manager  Details of 
--  Mode one 
select employee_id,last_name,salary,department_id
from employees e1
where exists(
    select * from employees e2 where e1.employee_id = e2.manager_id
          );
--  Mode two 
select distinct man.employee_id,man.last_name,man.salary,man.department_id
from employees emp
join employees man
on emp.manager_id = man.employee_id  ;

--  Mode three   Subquery 
select employee_id,last_name,salary,department_id
from employees
where employee_id in (
    select distinct manager_id from employees
    );

#13. Check the highest salary in each department   What is the minimum wage of the lowest department 
--  Mode one ：
select department_id,min(salary) from employees
where department_id = (
    select department_id from employees
    group by department_id
    having max(salary) = (
    select min(max_sal)
        from employees e, ( select max(salary) max_sal
                            from employees
                            where department_id is not null
                            group by department_id ) t_min_sal
    ));
--  Mode two ：
select department_id,min(salary) from employees
where department_id = (
    select department_id from employees
    group by department_id
    having max(salary) <= all (
        select max(salary) from employees
        group by department_id)
    );
--  Mode three ：
select min(salary) from employees
where department_id = (
    select department_id from employees
    group by department_id
    having max(salary) = (
       select max(salary) from employees
       group by department_id
       order by max(salary)
       limit 1 )
    );
--  Mode 4 ：
select min(salary) from employees e,( select department_id, max(salary) from employees
                                    group by department_id
                                    order by max(salary)
                                    limit 1) t_max_sal
where e.department_id = t_max_sal.department_id;
#14. Check the Department with the highest average wage  manager Details of : last_name,department_id,email,salary
--  Mode one ：
select last_name,department_id,email,salary
from employees
where employee_id = (
    select manager_id
    from departments
    where department_id = (
        select department_id from employees
        group by department_id
        order by avg(salary) desc
        limit 1)
    );
--  Mode two ：
select last_name,department_id,email,salary
from employees
where employee_id = (
select manager_id from departments
where department_id = (
    select department_id
    from employees
    group by department_id
    having avg(salary) >= all (
        select avg(salary)
        from employees
        where department_id is not null
        group by department_id)
    )
 );
--  Mode three ：
select last_name,department_id,email,salary from employees
where employee_id = (
    select manager_id from departments
    where department_id = (
        select department_id
        from employees
        group by department_id
        having avg(salary) = (
           select max(avg_sal) from (select avg(salary) avg_sal
           from employees
           where department_id is not null
           group by department_id) t_avg_sql
    ))
 );
#15． Query the department number of the Department , It does not include job_id yes "ST_CLERK" Your department number 
--  Mode one ：
select department_id
from departments
where department_id not in (
    select distinct department_id
    from employees e1
    where job_id = 'ST_CLERK'
    );
--  Mode two ：
select department_id
from departments d
where  not exists (
    select *
    from employees e1
    where d.department_id = e1.department_id and job_id = 'ST_CLERK'
    );
#16． Select all employees without managers last_name
select last_name
from employees e1
where not exists(
    select * from employees e2 where e1.manager_id = e2.employee_id
    );

#17． Check the employee number 、 full name 、 Employment time 、 Wages , The manager of the employee is  'De Haan '
--  Mode one ： Self join 
select e1.employee_id,e1.last_name,e1.hiredate,e1.salary
from employees e1,employees e2
where e1.manager_id = e2.employee_id and e2.last_name = 'De Haan';

--  Mode two ： Subquery 
select employee_id,last_name,hiredate,salary
from employees
where manager_id in (
    select employee_id from employees where last_name = 'De Haan'
    );

#18. Query the employee number of employees in each department whose salary is higher than the average salary of the Department , Name and salary （ Correlation subquery )
--  Mode one ：
select employee_id,salary,last_name
from employees e1
where salary > (
    select avg(salary) from employees e2 where e1.department_id = e2.department_id
    );
--  Mode two ：
select employee_id,salary,last_name
from employees e,(select department_id,avg(salary) avg_sal from employees group by department_id) t_avg_sal
where e.department_id = t_avg_sal.department_id and e.salary > t_avg_sal.avg_sal;

#19. Query the number of departments under each department is greater than 5 Department name of （ Correlation subquery )I
select department_name
from departments d
where 5 < (
    select count(*) from employees e where e.department_id = d.department_id
    );
#20. The number of departments in each country is greater than 2 Country number of ( Correlation subquery )
select country_id
from locations l
where 2 < (
    select count(*) from departments d where l.location_id = d.location_id
    );