SQL Exercises: Employees who earn second lowest salary of all
From the following table, write a SQL query to find those employees who earn the second-lowest salary of all the employees. Return all the fields 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 | 2003-06-17 | AD_PRES | 24000.00 | 0.00 | 0 | 90 | | 101 | Neena | Kochhar | NKOCHHAR | 515.123.4568 | 2005-09-21 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 102 | Lex | De Haan | LDEHAAN | 515.123.4569 | 2001-01-13 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 103 | Alexander | Hunold | AHUNOLD | 590.423.4567 | 2006-01-03 | IT_PROG | 9000.00 | 0.00 | 102 | 60 | | 104 | Bruce | Ernst | BERNST | 590.423.4568 | 2007-05-21 | IT_PROG | 6000.00 | 0.00 | 103 | 60 | | 105 | David | Austin | DAUSTIN | 590.423.4569 | 2005-06-25 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 106 | Valli | Pataballa | VPATABAL | 590.423.4560 | 2006-02-05 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 107 | Diana | Lorentz | DLORENTZ | 590.423.5567 | 2007-02-07 | IT_PROG | 4200.00 | 0.00 | 103 | 60 | | 108 | Nancy | Greenberg | NGREENBE | 515.124.4569 | 2002-08-17 | FI_MGR | 12008.00 | 0.00 | 101 | 100 | | 109 | Daniel | Faviet | DFAVIET | 515.124.4169 | 2002-08-16 | FI_ACCOUNT | 9000.00 | 0.00 | 108 | 100 | | 110 | John | Chen | JCHEN | 515.124.4269 | 2005-09-28 | FI_ACCOUNT | 8200.00 | 0.00 | 108 | 100 | .................... +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Sample Solution:
-- Selecting all columns (*) from the 'employees' table, aliased as 'm'
SELECT *
-- Filtering rows from the 'employees' table, aliased as 'm', based on the condition that the count of distinct 'salary' values is equal to 2
FROM employees m
-- Subquery to count the number of distinct 'salary' values where the 'salary' is less than or equal to the 'salary' in the outer query
WHERE 2 = (SELECT COUNT(DISTINCT salary )
FROM employees
WHERE salary <= m.salary);
Sample Output:
employee_id first_name last_name email phone_number hire_date job_id salary commission_pct manager_id department_id 128 Steven Markle SMARKLE 650.124.1434 2008-03-08ST_CLERK 2200.00 0.00 120 50 136 Hazel Philtanker HPHILTAN 650.127.1634 2008-02-06ST_CLERK 2200.00 0.00 122 50
Code Explanation:
The said query in SQL that selects all employees who have the second-lowest distinct salary in the employees table.
A subquery is used to count the number of distinct salaries that are less than or equal to each employee's salary in the outer query, based on the data in the subquery. Exactly two distinct salaries will be less than or equal to the second lowest salary of an employee. Therefore, the outer query selects all employees for whom the subquery returns a count of 2.
Alternative Solution:
Using a Self-Join:
SELECT m.*
FROM employees m
JOIN employees e ON e.salary <= m.salary
GROUP BY m.employee_id, m.salary
HAVING COUNT(DISTINCT e.salary) = 2;
Practice Online
Query Visualization:
Duration:
Rows:
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