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SQL Exercise: Employees who earn more than 100 as daily salary

SQL employee Database: Exercise-29 with Solution

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29. From the following table, write a SQL query to find out which employees earn more than 100 per day as a salary. Return employee ID, employee name, salary, and experience.

Sample table: employees


Pictorial Presentation:

SQL exercises on employee Database: List the id, name, salary, and experience of all the employees who earn more than 100 as daily salary

Sample Solution:

SELECT emp_id,
       emp_name,
       salary,
       age(CURRENT_DATE, hire_date) "Experience"
FROM employees
WHERE (salary/30)>100;

Sample Output:

 emp_id | emp_name | salary  |       Experience
--------+----------+---------+-------------------------
  68319 | KAYLING  | 6000.00 | 26 years 2 mons 12 days
  67858 | SCARLET  | 3100.00 | 20 years 9 mons 11 days
  69062 | FRANK    | 3100.00 | 26 years 1 mon 27 days
(3 rows)

Explanation:

The said query in SQL that selects the "emp_id", "emp_name", "salary", and experience (in years and months) from the 'employees' table and filters the results based on a calculation using the "salary" column.

The experience of the employee in years and months, calculated using the "age" function applied to the "hire_date" column.

The WHERE clause filters the results to include only those employees whose daily salary (calculated by dividing the monthly salary by 30) is greater than 100.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercise: Experiences of all employees working with Manger 68319.
Next SQL Exercise: Employees retiring after 31-Dec-99 after 8 years.

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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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