SQL Exercise: List the employees who are senior to their own manager
SQL employee Database: Exercise-63 with Solution
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63. From the following table, write a SQL query to find those employees who are senior to their manager. Return complete information about the employees.
Pictorial Presentation:

Sample table: employees
Sample Solution:
SELECT *
FROM employees w,
employees m
WHERE w.manager_id = m.emp_id
AND w.hire_date < m.hire_date;
Sample Output:
emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id | emp_id | emp_name | job_name | manager_id | hire_date | salary | commission | dep_id --------+----------+----------+------------+------------+---------+------------+--------+--------+----------+-----------+------------+------------+---------+------------+-------- 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 | 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 67832 | CLARE | MANAGER | 68319 | 1991-06-09 | 2550.00 | | 1001 | 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 65646 | JONAS | MANAGER | 68319 | 1991-04-02 | 2957.00 | | 2001 | 68319 | KAYLING | PRESIDENT | | 1991-11-18 | 6000.00 | | 1001 63679 | SANDRINE | CLERK | 69062 | 1990-12-18 | 900.00 | | 2001 | 69062 | FRANK | ANALYST | 65646 | 1991-12-03 | 3100.00 | | 2001 64989 | ADELYN | SALESMAN | 66928 | 1991-02-20 | 1700.00 | 400.00 | 3001 | 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 65271 | WADE | SALESMAN | 66928 | 1991-02-22 | 1350.00 | 600.00 | 3001 | 66928 | BLAZE | MANAGER | 68319 | 1991-05-01 | 2750.00 | | 3001 (6 rows)
Explanation:
The said query in SQL that selects all columns from the 'employees' table twice, with different aliases "w" and "m" to find all employees who were hired before their manager.
The query joins the 'employees' table to itself, to create a relationship between the employee and their manager based on the emp_id and manager_id columns, and the hire date comparison then filters the results to only show employees who were hired earlier than their manager using the WHERE clause.
Relational Algebra Expression:

Relational Algebra Tree:

Practice Online
Sample Database: employee

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Previous SQL Exercise: Employees location, grade, and experience over 25 years.
Next SQL Exercise: List the employee with their grade for the grade 4.
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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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