SQL Exercise: Employees under a given department ORDER BY dep_id ASC
SQL employee Database: Exercise-82 with Solution
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82. From the following table, write a SQL query to find the location of all the employees working in the FINANCE or AUDIT department. Sort the result-set in ascending order by department ID. Return complete information about the employees.
Pictorial Presentation:

Sample table: employees
Sample table: department
Sample Solution:
SELECT *
FROM employees e,
department d
WHERE (dep_name = 'FINANCE'
OR dep_name ='AUDIT')
AND e.dep_id = d.dep_id
ORDER BY e.dep_id ASC;
OR
SELECT *
FROM employees e,
department d
WHERE d.dep_name IN ('FINANCE',
'AUDIT')
AND e.dep_id = d.dep_id
ORDER BY e.dep_id ASC;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id | dep_id | dep_name | dep_location --------+----------+-----------+------------+------------+---------+------------+--------+--------+----------+-------------- 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 | 1001 | FINANCE | SYDNEY 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 | 1001 | FINANCE | SYDNEY 69324 | MARKER | CLERK | 67832 | 1992-01-23 | 1400.00 | | 1001 | 1001 | FINANCE | SYDNEY 67858 | SCARLET | ANALYST | 65646 | 1997-04-19 | 3100.00 | | 2001 | 2001 | AUDIT | MELBOURNE 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 | 2001 | AUDIT | MELBOURNE 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 | 2001 | AUDIT | MELBOURNE 68736 | ADNRES | CLERK | 67858 | 1997-05-23 | 1200.00 | | 2001 | 2001 | AUDIT | MELBOURNE 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 | 2001 | AUDIT | MELBOURNE (8 rows)
Explanation:
The said query in SQL that selects all columns (*) from the employees table and the department table where the department name is either 'FINANCE' or 'AUDIT', and the dep_id column in the employees table matches the dep_id column in the department table.
The WHERE clause includes only those employees who belong to the 'FINANCE' or 'AUDIT' and joins the employees and department tables on the dep_id column.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: Sort employees by designation, joining after 1991.
Next SQL Exercise: List the employees along with grades in ascending 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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