SQL Exercise: List the employees in given department, joined after 91

SQL employee Database: Exercise-65 with Solution

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65. From the following table, write a SQL query to find those employees who joined after 1991, excluding MARKER or ADELYN in the departments PRODUCTION or AUDIT. Return employee name.

Pictorial Presentation:

SQL exercises on employee Database: List the employees in department PRODUCTION or AUDIT who joined after 1991 and they are not MARKER or ADELYN to their name

Sample table: employees

Sample table: department

Sample table: salary_grade

Sample Solution:

SELECT e.emp_name
FROM employees e,
     department d,
     salary_grade s
WHERE e.dep_id = d.dep_id
  AND d.dep_name IN ('PRODUCTION',
  AND e.salary BETWEEN s.min_sal AND s.max_sal
  AND e.emp_name NOT IN ('MARKER',
  AND to_char(hire_date,'YYYY') >'1991';

Sample Output:

(2 rows)


The said query in SQL that retrieves the names of employees who belong to the "PRODUCTION" or "AUDIT" departments, have a salary within the range of their grade, were not hired before 1991, and have names other than "MARKER" and "ADELYN" from the 'employees', 'department', and 'salary_grade' tables.

The WHERE clause first joins the 'employees' and 'department' tables based on the department ID column.

It then retrieve employees who belong to the departments named "PRODUCTION" or "AUDIT" and then it checks whether the employee's salary falls within the range of minimum and maximum salaries defined for their grade in the 'salary_grade' table.

The WHERE clause then excludes employees with the names "MARKER" and "ADELYN" and includes only those employees who were hired after 1991 from the results.

Practice Online

Sample Database: employee

employee database structure

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Previous SQL Exercise: List the employee with their grade for the grade 4.
Next SQL Exercise: List the employees in ascending order of their salaries.

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SQL: Tips of the Day

Grouped LIMIT in PostgreSQL: Show the first N rows for each group?

db=# SELECT * FROM xxx;
 id | section_id | name
  1 |          1 | A
  2 |          1 | B
  3 |          1 | C
  4 |          1 | D
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
  8 |          2 | H
(8 rows)

I need the first 2 rows (ordered by name) for each section_id, i.e. a result similar to:

id | section_id | name
  1 |          1 | A
  2 |          1 | B
  5 |          2 | E
  6 |          2 | F
  7 |          3 | G
(5 rows)

PostgreSQL v9.3 you can do a lateral join

select distinct t_outer.section_id, t_top.id, t_top.name from t t_outer
join lateral (
    select * from t t_inner
    where t_inner.section_id = t_outer.section_id
    order by t_inner.name
    limit 2
) t_top on true
order by t_outer.section_id;

Database: PostgreSQL

Ref: https://bit.ly/3AfYwZI


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