MySQL Subquery Exercises: Find the employee ID, first and last name, and salary of all employees whose salary is above the average salary for their departments
MySQL Subquery: Exercise-14 with Solution
Write a MySQL query to display the employee ID, first name, last name, salary of all employees whose salary is above average for their departments.
Sample table: employees
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+ | EMPLOYEE_ID | FIRST_NAME | LAST_NAME | EMAIL | PHONE_NUMBER | HIRE_DATE | JOB_ID | SALARY | COMMISSION_PCT | MANAGER_ID | DEPARTMENT_ID | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+ | 100 | Steven | King | SKING | 515.123.4567 | 1987-06-17 | AD_PRES | 24000.00 | 0.00 | 0 | 90 | | 101 | Neena | Kochhar | NKOCHHAR | 515.123.4568 | 1987-06-18 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 102 | Lex | De Haan | LDEHAAN | 515.123.4569 | 1987-06-19 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 103 | Alexander | Hunold | AHUNOLD | 590.423.4567 | 1987-06-20 | IT_PROG | 9000.00 | 0.00 | 102 | 60 | | 104 | Bruce | Ernst | BERNST | 590.423.4568 | 1987-06-21 | IT_PROG | 6000.00 | 0.00 | 103 | 60 | | 105 | David | Austin | DAUSTIN | 590.423.4569 | 1987-06-22 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 106 | Valli | Pataballa | VPATABAL | 590.423.4560 | 1987-06-23 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 107 | Diana | Lorentz | DLORENTZ | 590.423.5567 | 1987-06-24 | IT_PROG | 4200.00 | 0.00 | 103 | 60 | | 108 | Nancy | Greenberg | NGREENBE | 515.124.4569 | 1987-06-25 | FI_MGR | 12000.00 | 0.00 | 101 | 100 | ......... | 206 | William | Gietz | WGIETZ | 515.123.8181 | 1987-10-01 | AC_ACCOUNT | 8300.00 | 0.00 | 205 | 110 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Sample table: departments
+---------------+----------------------+------------+-------------+ | DEPARTMENT_ID | DEPARTMENT_NAME | MANAGER_ID | LOCATION_ID | +---------------+----------------------+------------+-------------+ | 10 | Administration | 200 | 1700 | | 20 | Marketing | 201 | 1800 | | 30 | Purchasing | 114 | 1700 | | 40 | Human Resources | 203 | 2400 | | 50 | Shipping | 121 | 1500 | | 60 | IT | 103 | 1400 | | 70 | Public Relations | 204 | 2700 | | 80 | Sales | 145 | 2500 | | 90 | Executive | 100 | 1700 | | 100 | Finance | 108 | 1700 | .......... | 270 | Payroll | 0 | 1700 | +---------------+----------------------+------------+-------------+
Code:
-- Selecting the employee_id and first name of employees
SELECT employee_id, first_name
-- Selecting data from the employees table, aliasing it as 'A'
FROM employees AS A
-- Filtering the result set to include only employees whose salary is greater than the average salary of their department
WHERE salary >
-- Subquery to calculate the average salary for each department and compare with each employee's salary
(SELECT AVG(salary) FROM employees WHERE department_id = A.department_id);
Explanation:
- This MySQL code selects the employee_id and first name of employees from a table named "employees", aliasing it as 'A' for easier reference.
- It filters the results to only include employees whose salary is greater than the average salary of their department.
- This is achieved by using a subquery to calculate the average salary for each department (specified by the department_id) and comparing it with each employee's salary in the outer query.
- The subquery is correlated with the outer query by referencing the department_id of each employee in the WHERE clause.
MySQL Subquery Syntax:
- The subquery (inner query) executes once before the main query (outer query) executes.
- The main query (outer query) use the subquery result.
MySQL AVG() function calculates the average value of a set of values or an expression.
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