SQL Exercise: List id, name, department id, location of all employees

Last update on January 20 2026 12:30:51 (UTC/GMT +8 hours)

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93. From the following table, write a SQL query to list the employee id, name, department id, location 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 |
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+

View the table

Sample table: department

 departmentid |       name       | head
--------------+------------------+------
            1 | General Medicine |    4
            2 | Surgery          |    7
            3 | Psychiatry       |    9

Sample Solution:

SELECT e.emp_id,
       e.emp_name,
       e.dep_id,
       d.dep_location
FROM employees e,
     department d
WHERE e.dep_id = d.dep_id ;

Sample Output:

 emp_id | emp_name | dep_id | dep_location
--------+----------+--------+--------------
  68319 | KAYLING  |   1001 | SYDNEY
  66928 | BLAZE    |   3001 | PERTH
  67832 | CLARE    |   1001 | SYDNEY
  65646 | JONAS    |   2001 | MELBOURNE
  67858 | SCARLET  |   2001 | MELBOURNE
  69062 | FRANK    |   2001 | MELBOURNE
  63679 | SANDRINE |   2001 | MELBOURNE
  64989 | ADELYN   |   3001 | PERTH
  65271 | WADE     |   3001 | PERTH
  66564 | MADDEN   |   3001 | PERTH
  68454 | TUCKER   |   3001 | PERTH
  68736 | ADNRES   |   2001 | MELBOURNE
  69000 | JULIUS   |   3001 | PERTH
  69324 | MARKER   |   1001 | SYDNEY
(14 rows)

Explanation:

The said query in SQL that selects the emp_id, emp_name, dep_id from the employees table and the dep_location from the department table. The query joins the two tables on the dep_id column where they have the same values.

Relational Algebra Expression: