SQL Exercise: List employee id of all the departments 1001 and 2001

Last update on December 20 2024 12:00:11 (UTC/GMT +8 hours)

[An editor is available at the bottom of the page to write and execute the scripts.]

94. From the following table, write a SQL query to find those employees who work in the department ID 1001 or 2001. Return employee ID, employee name, department ID, department location, and department name.

Sample table: employees

Sample table: department

Sample Solution:

SELECT e.emp_id,
       e.emp_name,
       e.dep_id,
       d.dep_location,
       d.dep_name
FROM employees e,
     department d
WHERE e.dep_id = d.dep_id
  AND e.dep_id IN (1001,
                   2001);

Sample Output:

emp_id | emp_name | dep_id | dep_location | dep_name 
--------+----------+--------+--------------+----------
  68319 | KAYLING  |   1001 | SYDNEY       | FINANCE
  67832 | CLARE    |   1001 | SYDNEY       | FINANCE
  65646 | JONAS    |   2001 | MELBOURNE    | AUDIT
  68736 | ADNRES   |   2001 | MELBOURNE    | AUDIT
  69324 | MARKER   |   1001 | SYDNEY       | FINANCE
  67858 | SCARLET  |   2001 | MELBOURNE    | AUDIT
  69062 | FRANK    |   2001 | MELBOURNE    | AUDIT
  63679 | SANDRINE |   2001 | MELBOURNE    | AUDIT
(8 rows)

Explanation:

The said query in SQL that selects the emp_id, emp_name, dep_id, dep_location, and dep_name from the employees and department tables where the dep_id in the employees table is either 1001 or 2001.

The WHERE clause makes an implicit join between 'employees' and 'department' tables based on the dep_id column and includes only rows where the dep_id in the 'employees' table is either 1001 or 2001.

Relational Algebra Expression:

Relational Algebra Tree:

Practice Online

Sample Database: employee

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous SQL Exercise: List id, name, department id, location of all employees.
Next SQL Exercise: List employees salary within min_salary and max_salary.