SQL Exercise: Sort the employees by salary less than 1000
SQL employee Database: Exercise-73 with Solution
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73. From the following table, write a SQL query to find those employees who earn less than 1000. Sort the result-set in ascending order by salary. Return complete information about the employees.
Pictorial Presentation:

Sample table: employees
Sample Solution:
SELECT *
FROM employees
WHERE salary < 1000
ORDER BY salary;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+----------+------------+------------+--------+------------+-------- 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 (1 row)
Explanation:
The said query in SQL that returns all the rows and columns from the 'employees' table where the "salary" column is less than 1000, and sorts the result set in ascending order on "salary" column.
Relational Algebra Expression:

Relational Algebra Tree:

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Sample Database: employee

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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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