SQL Exercise: Senior employees with grades over 3 work under KAYLING
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25. From the following table, write a SQL query to find the most senior employee of grade 4 or 5, work under KAYLING. Return complete information about the 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 | ......... | 206 | William | Gietz | WGIETZ | 515.123.8181 | 2002-06-07 | AC_ACCOUNT | 8300.00 | 0.00 | 205 | 110 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Sample table: salary_grade
grade | min_sal | max_sal
-------+---------+---------
1 | 800 | 1300
2 | 1301 | 1500
3 | 1501 | 2100
4 | 2101 | 3100
5 | 3101 | 9999
Sample Solution:
SELECT *
FROM employees
WHERE hire_date IN
(SELECT min(hire_date)
FROM employees
WHERE emp_id IN
(SELECT emp_id
FROM employees e,
salary_grade s
WHERE e.salary BETWEEN s.min_sal AND s.max_sal
AND s.grade IN (4,
5)))
AND manager_id IN
(SELECT emp_id
FROM employees
WHERE emp_name = 'KAYLING');
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+----------+------------+------------+---------+------------+-------- 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 (1 row)
Explanation:
The said query in SQL that returns employees from the employees table who were hired on the earliest date and have a salary that falls within the range of salary grades 4 or 5 and have 'KAYLING' as their manager.
The subquery selects the "emp_id" of employees who have a salary that falls within the range of salary grades 4 or 5. It does this by joining the 'employees' table with the 'salary_grade' table on the condition that the "salary" of an employee falls within the "min_sal" and "max_sal" range of the corresponding salary grade. It then selects the minimum hire date of these employees.
The outer query then selects all rows from the 'employees' table where the "hire_date" is equal to the value returned by the subquery and their "manager_id" is the "emp_id" of an employee whose name is 'KAYLING'.
Go to:
PREV : Employees who joined in 1991 as the most senior of 1991.
NEXT : Find the total salary given to the MANAGER.
Practice Online
Sample Database: employees
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