MySQL Subquery Exercises: Query to get nth maximum salaries of employees
MySQL Subquery: Exercise-22 with Solution
Write a MySQL query to get nth maximum salaries of employees.
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 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Code:
-- Selecting all columns from the employees table, aliasing it as 'emp1'
SELECT *
-- Selecting data from the employees table, aliasing it as 'emp1'
FROM employees emp1
-- Filtering the result set to include only employees where the count of distinct salaries greater than the salary of the current employee is 1
WHERE (1) = (
-- Subquery to count the number of distinct salaries greater than the salary of each employee
SELECT COUNT(DISTINCT(emp2.salary))
-- Selecting data from the employees table, aliasing it as 'emp2'
FROM employees emp2
-- Filtering the result set to include only distinct salaries greater than the salary of the employee in the outer query (emp1.salary)
WHERE emp2.salary > emp1.salary
);
Explanation:
- This MySQL code selects all columns from the "employees" table and aliases it as 'emp1'.
- It filters the result set to include only employees where the count of distinct salaries greater than the salary of the current employee is 1.
- This is achieved using a subquery where the number of distinct salaries greater than the salary of each employee is counted. The outer query compares this count with the constant value 1.
- If there's exactly one distinct salary greater than the salary of an employee, that employee will be included in the result set.
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.
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Structure of 'hr' database :
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