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:

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

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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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