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

SQL employee Database: Exercise-93 with Solution

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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


Sample table: department


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:

Relational Algebra Expression: List the employee id, name, department id, location of all the employees.

Relational Algebra Tree:

Relational Algebra Tree: List the employee id, name, department id, location of all the employees.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Compute the total salary for each job in 1991.
Next SQL Exercise: List employee id of all the departments 1001 and 2001.

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