SQL Exercise: List the employees not in department 2001
SQL employee Database: Exercise-11 with Solution
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11. From the following table, write a SQL query to find those employees who do not belong to the department 2001. Return complete information about the employees.
Sample table: employees
Pictorial Presentation:

Sample Solution:
SELECT *
FROM employees
WHERE dep_id NOT IN (2001);
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 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 64989 | ADELYN | SALESMAN | 66928 | 1991-02-20 | 1700.00 | 400.00 | 3001 65271 | WADE | SALESMAN | 66928 | 1991-02-22 | 1350.00 | 600.00 | 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 69000 | JULIUS | CLERK | 66928 | 1991-12-03 | 1050.00 | | 3001 69324 | MARKER | CLERK | 67832 | 1992-01-23 | 1400.00 | | 1001 (9 rows)
Explanation:
The said query in SQL that selects all columns from the 'employees' table where the dep_id is not equal to 2001.
The "WHERE" clause filters the results for all employees whose department ID does not match the value 2001
The "NOT IN" operator excludes those department that matched with the values of dep_id 2001.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: Display the unique department with jobs.
Next SQL Exercise: List the employees who joined before 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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