SQL Exercise: Employees whose salary is greater than their managers
SQL employee Database: Exercise-54 with Solution
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54. From the following table, write a SQL query to identify employees whose salaries are higher than their managers' salaries. Return employee name, job name, manager ID, salary, manager name, manager's salary.
Sample table: employees
Pictorial Presentation:

Sample Solution:
SELECT w.emp_name,
w.job_name,
w.manager_id,
w.salary,
m.emp_name "Manager",
m.emp_id,
m.salary "Manager_Salary"
FROM employees w,
employees m
WHERE w.manager_id = m.emp_id
AND w.salary > m.salary;
Sample Output:
emp_name | job_name | manager_id | salary | Manager | emp_id | Manager_Salary ----------+----------+------------+---------+---------+--------+---------------- SCARLET | ANALYST | 65646 | 3100.00 | JONAS | 65646 | 2957.00 FRANK | ANALYST | 65646 | 3100.00 | JONAS | 65646 | 2957.00 (2 rows)
Explanation:
The given query in SQL that selects the employee name, job name, manager ID, salary, manager name aliased as "Manager", manager ID, and manager salary aliased as "Manager_Salary" for all employees from the employees table.
The query is using a self-join to combine data from the employees table with itself based on the manager_id column.
The WHERE clause filters those employees whose salary is greater than their manager's salary.
Practice Online
Sample Database: employee

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Previous SQL Exercise: Employees earning 60000 or not working as analysts.
Next SQL Exercise: Employees whose salary is in a range and location PERTH.
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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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