Database Mysql Basics Everyone Should Know

SQL Queries and Database Basics

Here we are going to understand database basics, starting with creating simple tables.

Create the following tables

Queries Based on tables

Simple Queries:

     Prob: List all the employee details.

     Sol : SQL > Select * from employee;

     Prob: List all the department details

     Sol: SQL > Select * from department;

     Prob: List all job details

     Sol: SQL > Select * from job;

     Prob: List all the locations

     Sol: SQL > Select * from loc;

     Prob: List out first name,last name,salary, commission for all employees

     Sol: SQL > Select first_name, last_name, salary, commission from employee;

     Prob: List out employee_id,last name,department id for all  employees and rename    

     employee id as “ID  of the employee”, last name as “Name of the employee”,

     department id as  “department  ID”

     Sol: SQL > Select employee_id “id of the employee”, last_name “name", department 

     id as “department id” from employee;

     Prob: List out the employees anuual salary with their names only.

     Sol: SQL > Select last_name, salary*12 “annual salary” from employee

Where Conditions:

    Prob: List the details about “SMITH”

    Sol:  SQL > Select * from employee where last_name=’SMITH’;

    Prob: List out the employees who are working in department 20

    Sol: SQL > Select * from employee where department_id=20

    Prob: List out the employees who are earning salary between 3000 and 4500

    Sol: SQL > Select * from employee where salary between 3000 and 4500

    Prob: List out the employees who are working in department 10 or 20

    Sol: SQL > Select * from employee where department_id in (20,30)

    Prob: Find out the employees who are not working in department 10 or 30

    Sol: SQL > Select last_name, salary, commission, department_id from employee where 

    department_id not in (10,30)

    Prob: List out the employees whose name starts with “S”

    Sol: SQL > Select * from employee where last_name like ‘S%’

    Prob:List out the employees whose name start with “S” and end with “H”

    Sol: SQL > Select * from employee where last_name like ‘S%H’

    Prob: List out the employees whose name length is 4 and start with “S”

    Sol: SQL > Select * from employee where last_name like ‘S___’

    Prob: List out the employees who are working in department 10 and draw the salaries more than 3500

    Sol:SQL > Select * from employee where department_id=10 and salary>3500

    Prob:list out the employees who are not receiving commission.

    Sol: SQL > Select * from employee where commission is Null

Order By Clause:

    Prob:List out the employee id, last name in ascending order based on the employee id.

    Sol: SQL > Select employee_id, last_name from employee order by employee_id

    Prob:List out the employee id, name in descending order based on salary column

    Sol: SQL > Select employee_id, last_name, salary from employee order by salary desc

    Prob:list out the employee details according to their last_name in ascending order and   

    salaries in descending order

    Sol: SQL > Select employee_id, last_name, salary from employee order by last_name,

    salary desc

    Prob:list out the employee details according to their last_name in ascending order and 

    then on department_id in descending order.

    Sol: SQL > Select employee_id, last_name, salary from employee order by last_name, 

   department_id desc

    

Group By & Having Clause:

   Prob: How many employees who are working in different departments wise in the 

   organization

   Sol:  SQL > Select department_id, count(*), from employee group by department_id

   Prob: List out the department wise maximum salary, minimum salary, average salary of 

   the employees

   Sol:  SQL > Select department_id, count(*), max(salary), min(salary), avg(salary) from 

   employee group by department_id

   Prob: List out the job wise maximum salary, minimum salary, average salaries of the 

   employees.

   Sol: SQL > Select job_id, count(*), max(salary), min(salary), avg(salary) from 

   employee group by job_id

   Prob: List out the no.of employees joined in every month in ascending order.

   Sol: SQL > Select to_char(hire_date,’month’)month, count(*) from employee group by 

   to_char(hire_date,’month’) order by month

   Prob: List out the no.of employees for each month and year, in the ascending order 

   based on the year, month.

   Sol: SQL > Select to_char(hire_date,’yyyy’) Year, to_char(hire_date,’mon’) Month,

   count(*) “No. of employees” from employee group by to_char(hire_date,’yyyy’),

   o_char(hire_date,’mon’)

   Prob: List out the department id having atleast four employees.

   Sol: SQL > Select department_id, count(*) from employee group by department_id having count(*)>=4

   Prob: How many employees in January month.

  Sol : SQL > Select to_char(hire_date,’mon’) month, count(*) from employee group by 
  to_char(hire_date,’mon’) having to_char(hire_date,’mon’)=’jan’

   Prob: How many employees who are joined in January or September month.

  Sol: SQL > Select to_char(hire_date,’mon’) month, count(*) from employee group by 
   to_char(hire_date,’mon’) having to_char(hire_date,’mon’) in (‘jan’,’sep’)

  Prob: How many employees who are joined in 1985.

  Sol : SQL > Select to_char(hire_date,’yyyy’) Year, count(*) from employee group by
  to_char(hire_date,’yyyy’) having to_char(hire_date,’yyyy’)=1985

  Prob: How many employees joined each month in 1985.

  Sol: SQL > Select to_char(hire_date,’yyyy’)Year, to_char(hire_date,’mon’) Month, count(*) “No.  of employees” from employee where to_char(hire_date,’yyyy’)=1985 group by  to_char(hire_date,’yyyy’),to_char(hire_date,’mon’)

  Prob:  How many employees who are joined in March 1985.

  Sol: SQL > Select to_char(hire_date,’yyyy’)Year, to_char(hire_date,’mon’) Month,
  count(*) “No. of employees” from employee where to_char(hire_date,’yyyy’)=1985 and
  to_char(hire_date,’mon’)=’mar’ group by  to_char(hire_date,’yyyy’),
  to_char(hire_date,’mon’)

  Prob: Which is the department id, having greater than or equal to 3 employees joined in
   April 1985.

  Sol: SQL > Select department_id, count(*) “No. of employees” from employee where 
  to_char(hire_date,’yyyy’)=1985 and to_char(hire_date,’mon’)=’apr’ group by
  to_char(hire_date,’yyyy’), to_char(hire_date,’mon’), department_id having
  count(*)>=3

Sub-Queries

   Prob: Display the employee who got the maximum salary.

   Sol: SQL > Select * from employee where salary=(select max(salary) from employee)

   Prob: Display the employees who are working in Sales department

  Sol : SQL > Select * from employee where department_id IN (select department_id from

  department where name=’SALES’)

  Prob:Display the employees who are working as “Clerk”.

  Sol: SQL > Select * from employee where job_id in (select job_id from job where

  function=’CLERK’

  Prob: Display the employees who are working in “New York”

  Sol: SQL > Select * from employee where department_id=(select department_id from

  department where location_id=(select location_id from location where regional_group=

  ’New York’))    

   Prob: Find out no.of employees working in “Sales” department.

   Sol: SQL > Select * from employee where department_id=(select department_id from

   department where name=’SALES’ group by department_id)

   Prob: Update the employees salaries, who are working as Clerk on the basis of 10%.

   Sol: SQL > Update employee set salary=salary*10/100 wehre job_id=(select job_id from job where function=’CLERK’)

   Prob: Delete the employees who are working in accounting department.

   Sol : SQL > delete from employee where department_id=(select department_id from department    where name=’ACCOUNTING’)

   Prob:  Display the second highest salary drawing employee details.

   Sol:  SQL > Select * from employee where salary=(select max(salary) from employee where salary <(select max(salary) from employee))

   Prob: Display the Nth highest salary drawing employee details.

   Sol: SQL > Select distinct e.salary from employee where & no-1=(select count(distinct salary)  from employee where sal>e.salary)

Sub-Query operators: (ALL,ANY,SOME,EXISTS)

   Prob: List out the employees who earn more than every employee in department 30.

   Sol: SQL > Select * from employee where salary > all (Select salary from employee where department_id=30)

   Prob: List out the employees who earn more than the lowest salary in department 30.

   Sol: SQL > Select * from employee where salary > any (Select salary from employee where department_id=30)

   Prob: Find out whose department has not employees.

   Sol: SQL > Select employee_id, last_name, department_id from employee e where not exists (select department_id from department d where d.department_id=e.department_id)

   Prob: Find out which department does not have any employees.

   Sol: SQL > Select name from department d where not exists (select last_name from employee e where d.department_id=e.department_id)

Co-Related Sub Queries:

   Prob: find out the employee who can earn greater than the average salary for their department.

  Sol: SQL > Select employee_id, last_name, salary, department_id from employee e where   salary > (select avg(salary) from employee where department_id=e.department_id)

Joins

Simple join

   Prob: List our employees with their department names

   Sol:  SQL > Select employee_id, last_name, name from employee e, department d where  e.department_id=d.department_id

   Prob: Display employees with their designations (jobs)

   Sol: SQL > Select employee_id, last_name, function from employee e, job j where 

   e.job_id=j.job_id

   Prob: Display the employees with their department name and regional groups

   Sol: SQL > Select employee_id, last_name, name, regional_group from employee e, department d, location l where e.department_id=d.department_id and d.location_id=l.location_id

   Prob: How many employees who are working in different departments and display with

   department name. 

   Sol: SQL > Select name, count(*) from employee e, department d where 

   d.department_id=e.department_id group by name

   Prob: How many employees who are working in sales department.

   Sol:  SQL > Select name, count(*) from employee e, department d where

   d.department_id=e.department_id group by name having name=’SALES’

   Prob: which is the department having greater than or equal to 5 employees and display the department name in ascending order.

   Sol:  SQL > Select name, count(*) from employee e, department d where 

   d.department_id=e.department_id group by name having count (*)>=5 order by name

   Prob: How many jobs in the organization with designations.

   Sol:  SQL > Select function, count(*) from employee e, job j where j.job_id=e.job_id group by    function

   Prob: How many employees working in “New York”.

   Sol: SQL > Select regional_group, count(*) from employee e, department d, location l where  e.department_id=d.department_id and d.location_id=l.location_id and regional_group=’NEW YORK’ group by regional_group

Non – Equi Join:

    Prob: Display employee details with salary grades.

   Sol: SQL > Select employee_id, last_name, grade_id from employee e, salary_grade s where salary between lower_bound and upper_bound order by last_name

   Prob: List out the no. of employees on grade wise.

   Sol : SQL > Select grade_id, count(*) from employee e, salary_grade s where salary between  lower_bound and upper_bound group by grade_id order by grade_id desc

   Prob: Display the employee salary grades and no of employees between 2000 and 5000 of salary.

   Sol: SQL > Select grade_id, count(*) from employee e, salary_grade s where salary between lower_bound and upper_bound and lower_bound>=2000 and lower_bound<=5000 group by grade_id order by grade_id desc

Self Join:

    Prob: Display the employee details with their manager names.

   Sol: SQL > Select e.last_name emp_name, m.last_name, mgr_name from employee e, employee m where e.manager_id=m.employee_id

   Prob: Display the employee details who earn more than their managers salaries.

   Sol: SQL > Select e.last_name emp_name, e.salary emp_salary, m.last_name, mgr_name,  m.salary mgr_salary from employee e, employee m where e.manager_id=m.employee_id and m.salary

   Prob: Show the no. of employees working under every manager.

   Sol: SQL > Select m.manager_id, count(*) from employee e, employee m where 

   e.employee_id=m.manager_id group by m.manager_id

Outer Join:

   Prob: Display employee details with all departments.

  Sol: SQL > Select last_name, d.department_id, d.name from employee e, department d where e.department_id(+)=d.department_id

  Prob: Display all employees in sales or operation departments.

  Sol : SQL > Select last_name, d.department_id, d.name from employee e, department d where e.department_id(+)=d.department_id and d.department_idin (select department_id from department where name IN (‘SALES’,’OPERATIONS’))

Set Operators:

Prob: List out the distinct jobs in Sales and Accounting Departments.

Sol : SQL > Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name=’SALES’)) union Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name=’ACCOUNTING’))

Prob: List out the ALL jobs in Sales and Accounting Departments.

Sol:  SQL > Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name=’SALES’)) union all Select  function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name=’ACCOUNTING’))

Prob: List out the common jobs in research and accounting departments in ascending orders.

Sol: SQL > Select function from job where job_id in (Select job_id from employee where department_id=(select department_id from department where name=’RESEARCH’)) intersect  Select function from job where job_id in (Select job_id from employee where department_id= (select department_id from department where name=’ACCOUNTING’)) order by function