MySQL Subquery Exercises: Find the name of the employees who are managers
MySQL Subquery: Exercise-4 with Solution
Write a MySQL query to find the name (first_name, last_name) of the employees who are managers.
Sample table: employees
+-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+ | EMPLOYEE_ID | FIRST_NAME | LAST_NAME | EMAIL | PHONE_NUMBER | HIRE_DATE | JOB_ID | SALARY | COMMISSION_PCT | MANAGER_ID | DEPARTMENT_ID | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+ | 100 | Steven | King | SKING | 515.123.4567 | 1987-06-17 | AD_PRES | 24000.00 | 0.00 | 0 | 90 | | 101 | Neena | Kochhar | NKOCHHAR | 515.123.4568 | 1987-06-18 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 102 | Lex | De Haan | LDEHAAN | 515.123.4569 | 1987-06-19 | AD_VP | 17000.00 | 0.00 | 100 | 90 | | 103 | Alexander | Hunold | AHUNOLD | 590.423.4567 | 1987-06-20 | IT_PROG | 9000.00 | 0.00 | 102 | 60 | | 104 | Bruce | Ernst | BERNST | 590.423.4568 | 1987-06-21 | IT_PROG | 6000.00 | 0.00 | 103 | 60 | | 105 | David | Austin | DAUSTIN | 590.423.4569 | 1987-06-22 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 106 | Valli | Pataballa | VPATABAL | 590.423.4560 | 1987-06-23 | IT_PROG | 4800.00 | 0.00 | 103 | 60 | | 107 | Diana | Lorentz | DLORENTZ | 590.423.5567 | 1987-06-24 | IT_PROG | 4200.00 | 0.00 | 103 | 60 | | 108 | Nancy | Greenberg | NGREENBE | 515.124.4569 | 1987-06-25 | FI_MGR | 12000.00 | 0.00 | 101 | 100 | ......... | 206 | William | Gietz | WGIETZ | 515.123.8181 | 1987-10-01 | AC_ACCOUNT | 8300.00 | 0.00 | 205 | 110 | +-------------+-------------+-------------+----------+--------------------+------------+------------+----------+----------------+------------+---------------+
Code:
-- Selecting the first name and last name of employees
SELECT first_name, last_name
-- Selecting data from the employees table
FROM employees
-- Filtering the result set to include only employees whose employee_id is in the set of manager_ids
WHERE (employee_id IN
-- Subquery to select manager_ids from the employees table
(SELECT manager_id FROM employees)
);
Explanation:
- This MySQL code selects the first name and last name of employees from a table named "employees".
- It filters the results to only include employees whose employee_id is in the set of manager_ids.
- This is achieved by using a subquery to select manager_ids from the "employees" table, and then comparing each employee's employee_id with the manager_ids obtained from the subquery.
MySQL Subquery Syntax:
operand comparison_operator operand IN (subquery) operand comparison_operator SOME (subquery)
Where comparison_operator is one of these operators
= > < >= <= <> !=
and IN operator checks whether a value is within a set of values.
For example :
mysql> SELECT 2 IN (0,3,5,7);
-> 0
mysql> SELECT 'wefwf' IN ('wee','wefwf','weg');
-> 1
When used with a subquery, the word IN is an alias for = ANY. Thus, these two statements are the same:
SELECT s1 FROM t1 WHERE s1 = ANY (SELECT s1 FROM t2);
SELECT s1 FROM t1 WHERE s1 IN (SELECT s1 FROM t2);
IN and = ANY are not synonyms when used with an expression list. IN can take an expression list, but = ANY cannot.
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Structure of 'hr' database :
MySQL Code Editor:
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