SQL Exercise: Employees in ascending and descending order
SQL employee Database: Exercise-67 with Solution
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67. From the following table, write a SQL query to list employees in ascending order on department ID and descending order on jobs. Return complete information about the employees.
Pictorial Presentation:
Sample table: employees
Sample Solution:
SELECT *
FROM employees
ORDER BY dep_id ASC,
job_name DESC;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+-----------+------------+------------+---------+------------+-------- 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 69324 | MARKER | CLERK | 67832 | 1992-01-23 | 1400.00 | | 1001 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 68736 | ADNRES | CLERK | 67858 | 1997-05-23 | 1200.00 | | 2001 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 67858 | SCARLET | ANALYST | 65646 | 1997-04-19 | 3100.00 | | 2001 65271 | WADE | SALESMAN | 66928 | 1991-02-22 | 1350.00 | 600.00 | 3001 64989 | ADELYN | SALESMAN | 66928 | 1991-02-20 | 1700.00 | 400.00 | 3001 66564 | MADDEN | SALESMAN | 66928 | 1991-09-28 | 1350.00 | 1500.00 | 3001 68454 | TUCKER | SALESMAN | 66928 | 1991-09-08 | 1600.00 | 0.00 | 3001 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 69000 | JULIUS | CLERK | 66928 | 1991-12-03 | 1050.00 | | 3001 (14 rows)
Explanation:
The said query in SQL that retrieves all the rows from the 'employees' table and order them first by the "dep_id" column in ascending order, and then by the "job_name" column in descending order.
This means that the employee records will be grouped first by department ID in ascending order, and within each department, the records will be ordered by job name in descending order.
Relational Algebra Expression:
Relational Algebra Tree:
Practice Online
Sample Database: employee
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Previous SQL Exercise: List the employees in ascending order of their salaries.
Next SQL Exercise: Display all the unique job in descending order.
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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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