SQL Exercise: List employees id, salary, and commission
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9. From the following table, write a SQL query to find the employee ID, salary, and commission of all 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 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Pictorial Presentation:
Sample Solution:
SELECT emp_id,
salary,
commission
FROM employees;
Sample Output:
emp_id | salary | commission --------+---------+------------ 68319 | 6000.00 | 66928 | 2750.00 | 67832 | 2550.00 | 65646 | 2957.00 | 67858 | 3100.00 | 69062 | 3100.00 | 63679 | 900.00 | 64989 | 1700.00 | 400.00 65271 | 1350.00 | 600.00 66564 | 1350.00 | 1500.00 68454 | 1600.00 | 0.00 68736 | 1200.00 | 69000 | 1050.00 | 69324 | 1400.00 | (14 rows)
Explanation:
The given query in SQL that selects the emp_id, salary, and commission columns from the 'employees' table.
Relational Algebra Expression:
Relational Algebra Tree:
Go to:
PREV : Without the spaces, count each employee name characters.
NEXT : Display the unique department with jobs.
Practice Online
Sample Database: employees
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