SQL Exercise: Find employees whose managers are hold given IDs
SQL SORTING and FILTERING on HR Database: Exercise-16 with Solution
16. From the following table, write a SQL query to find the employees whose managers hold the ID 120, 103, or 145. Return first name, last name, email, salary and manager ID.
Sample table : employees
Sample Solution:
SELECT first_name, last_name, email,
salary, manager_id
FROM employees
WHERE manager_id IN (120 , 103 , 145);
Sample Output:
first_name | last_name | email | salary | manager_id -------------+-------------+----------+----------+------------ Bruce | Ernst | BERNST | 6000.00 | 103 David | Austin | DAUSTIN | 4800.00 | 103 Valli | Pataballa | VPATABAL | 4800.00 | 103 Diana | Lorentz | DLORENTZ | 4200.00 | 103 Julia | Nayer | JNAYER | 3200.00 | 120 Irene | Mikkilineni | IMIKKILI | 2700.00 | 120 James | Landry | JLANDRY | 2400.00 | 120 Steven | Markle | SMARKLE | 2200.00 | 120 Peter | Tucker | PTUCKER | 10000.00 | 145 David | Bernstein | DBERNSTE | 9500.00 | 145 Peter | Hall | PHALL | 9000.00 | 145 Christopher | Olsen | COLSEN | 8000.00 | 145 Nanette | Cambrault | NCAMBRAU | 7500.00 | 145 Oliver | Tuvault | OTUVAULT | 7000.00 | 145 Winston | Taylor | WTAYLOR | 3200.00 | 120 Jean | Fleaur | JFLEAUR | 3100.00 | 120 Martha | Sullivan | MSULLIVA | 2500.00 | 120 Girard | Geoni | GGEONI | 2800.00 | 120 (18 rows)
Code Explanation:
The said query in SQL that retrieves the first name, last name, email, salary, and manager ID columns from the 'employees' table where the value in the "manager_id" column is one of the following values: 120, 103, or 145.
Relational Algebra Expression:
Relational Algebra Tree:
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Previous SQL Exercise: Find those employees hired before June 21st, 2002.
Next SQL Exercise: Employees have the given letters in their first name.
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SQL: Tips of the Day
Difference between natural join and inner join
One significant difference between INNER JOIN and NATURAL JOIN is the number of columns returned-
Consider:
TableA TableB +------------+----------+ +--------------------+ |Column1 | Column2 | |Column1 | Column3 | +-----------------------+ +--------------------+ | 1 | 2 | | 1 | 3 | +------------+----------+ +---------+----------+
The INNER JOIN of TableA and TableB on Column1 will return
SELECT * FROM TableA AS a INNER JOIN TableB AS b USING (Column1); SELECT * FROM TableA AS a INNER JOIN TableB AS b ON a.Column1 = b.Column1;
+------------+-----------+---------------------+ | a.Column1 | a.Column2 | b.Column1| b.Column3| +------------------------+---------------------+ | 1 | 2 | 1 | 3 | +------------+-----------+----------+----------+
The NATURAL JOIN of TableA and TableB on Column1 will return:
SELECT * FROM TableA NATURAL JOIN TableB +------------+----------+----------+ |Column1 | Column2 | Column3 | +-----------------------+----------+ | 1 | 2 | 3 | +------------+----------+----------+
Ref: https://bit.ly/3AG5CId
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