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SQL Exercise: List employees salary within min_salary and max_salary

SQL employee Database: Exercise-95 with Solution

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95. From the following table, write a SQL query to find those employees whose salary is in the range of minimum and maximum salary (Begin and end values are included.). Return employee ID, name, salary and grade.

Sample table: employees


Sample table: salary_grade


Sample Solution:

SELECT e.emp_id,
       e.emp_name,
       e.salary,
       s.grade
FROM employees e,
     salary_grade s
WHERE e.salary BETWEEN s.min_sal AND s.max_sal ;

Sample Output:

 emp_id | emp_name | salary  | grade
--------+----------+---------+-------
  63679 | SANDRINE |  900.00 |     1
  68736 | ADNRES   | 1200.00 |     1
  69000 | JULIUS   | 1050.00 |     1
  65271 | WADE     | 1350.00 |     2
  66564 | MADDEN   | 1350.00 |     2
  69324 | MARKER   | 1400.00 |     2
  64989 | ADELYN   | 1700.00 |     3
  68454 | TUCKER   | 1600.00 |     3
  66928 | BLAZE    | 2750.00 |     4
  67832 | CLARE    | 2550.00 |     4
  65646 | JONAS    | 2957.00 |     4
  67858 | SCARLET  | 3100.00 |     4
  69062 | FRANK    | 3100.00 |     4
  68319 | KAYLING  | 6000.00 |     5
(14 rows)

Explanation:

The said query in SQL that selects the employee ID, employee name, salary, and grade from the 'employees' and 'salary_grade' tables where salary falls within the minimum and maximum salary ranges in the 'salary_grade' table.

An implicit cross join builds between the tables 'employees' and 'salary_grade' which means that each row in the 'employees' table is combined with every row in the 'salary_grade' table and the WHERE clause filters the results to include only those employees whose salary falls within the minimum and maximum salary ranges in the 'salary_grade' table.

Relational Algebra Expression:

Relational Algebra Expression: List the employee id, name, salary, grade of all the employees whose salary within min_salary and max_salary.

Relational Algebra Tree:

Relational Algebra Tree: List the employee id, name, salary, grade of all the employees whose salary within min_salary and max_salary.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: List employee id of all the departments 1001 and 2001.
Next SQL Exercise: In ASC order, list managers and count their employees.

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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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