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SQL Exercise: List employees in ascending order on their experiences

SQL employee Database: Exercise-80 with Solution

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80. From the following table, write a SQL query to list the employee ID, name, hire date, current date and experience of the employees in ascending order on their experiences.

Pictorial Presentation:

SQL exercises on employee Database: List the employee id, name, hire_date, current date and experience of the employees in ascending order on their experiences

Sample table: employees


Sample Solution:

SELECT emp_id,
       emp_name,
       hire_date,
       CURRENT_DATE,
       age(CURRENT_DATE, hire_date) EXP
FROM employees
ORDER BY EXP ASC;

Sample Output:

 emp_id | emp_name | hire_date  |    date    |           exp
--------+----------+------------+------------+-------------------------
  68736 | ADNRES   | 1997-05-23 | 2018-02-01 | 20 years 8 mons 9 days
  67858 | SCARLET  | 1997-04-19 | 2018-02-01 | 20 years 9 mons 12 days
  69324 | MARKER   | 1992-01-23 | 2018-02-01 | 26 years 9 days
  69062 | FRANK    | 1991-12-03 | 2018-02-01 | 26 years 1 mon 29 days
  69000 | JULIUS   | 1991-12-03 | 2018-02-01 | 26 years 1 mon 29 days
  68319 | KAYLING  | 1991-11-18 | 2018-02-01 | 26 years 2 mons 13 days
  66564 | MADDEN   | 1991-09-28 | 2018-02-01 | 26 years 4 mons 3 days
  68454 | TUCKER   | 1991-09-08 | 2018-02-01 | 26 years 4 mons 23 days
  67832 | CLARE    | 1991-06-09 | 2018-02-01 | 26 years 7 mons 22 days
  66928 | BLAZE    | 1991-05-01 | 2018-02-01 | 26 years 9 mons
  65646 | JONAS    | 1991-04-02 | 2018-02-01 | 26 years 9 mons 29 days
  65271 | WADE     | 1991-02-22 | 2018-02-01 | 26 years 11 mons 7 days
  64989 | ADELYN   | 1991-02-20 | 2018-02-01 | 26 years 11 mons 9 days
  63679 | SANDRINE | 1990-12-18 | 2018-02-01 | 27 years 1 mon 14 days
(14 rows)

Explanation:

The said query in SQL that selects the emp_id, emp_name, hire_date, CURRENT_DATE, and EXP columns for each employee in the employees table, sorted in ascending order by their years of experience.

The age() function calculates the difference in years between the hire_date and CURRENT_DATE and the resulting column is aliased as EXP.

Practice Online


Sample Database: employee

employee database structure

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Previous SQL Exercises: Employees work as SALESMEN, sorted by their salary.
Next SQL Exercise: Sort employees by designation, joining after 1991.

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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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