SQL Exercise: List the employee who are not working under a manager
SQL employee Database: Exercise-26 with Solution
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26. From the following table, write a SQL query identify the employees who do not report to a manager. Return employee name, job name.
Sample table: employees
SELECT e.emp_name, e.job_name FROM employees e WHERE manager_id IS NULL;
emp_name | job_name --------------+-------------- KAYLING | PRESIDENT
The said query in SQL that retrieves a list of all employees who do not have a manager, along with their name and job position from the 'employees' table.
The WHERE clause that filters the results so that, it only includes the employees who do not have a value in the "manager_id" column, which indicates whether or not the "manager_id" column is null.
Relational Algebra Expression:
Relational Algebra Tree:
Sample Database: employee
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Previous SQL Exercise: List the employees who joined before 1 April 1991.
Next SQL Exercise: List all the employees joined on 1st may 91.
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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;
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