SQL Challenges-1: Duplicate Emails
SQL Challenges-1: Exercise-5 with Solution
From the following table, write a SQL query to find all the duplicate emails (no upper case letters) of the employees. Return email ids.
Input:
Table: employees
Structure:
Field | Type | Null | Key | Default | Extra |
---|---|---|---|---|---|
employee_id | int(11) | YES | |||
employee_name | varchar(255) | YES | |||
email_id | varchar(255) | YES |
Data:
employee_id | employee_name | email_id |
---|---|---|
101 | Liam Alton | [email protected] |
102 | Josh Day | [email protected] |
103 | Sean Mann | [email protected] |
104 | Evan Blake | [email protected] |
105 | Toby Scott | [email protected] |
Sample Solution:
SQL Code(MySQL):
CREATE TABLE IF NOT EXISTS employees(employee_id int, employee_name varchar(255), email_id varchar(255));
TRUNCATE TABLE employees;
INSERT INTO employees (employee_id,employee_name, email_id) VALUES ('101','Liam Alton', '[email protected]');
INSERT INTO employees (employee_id,employee_name, email_id) VALUES ('102','Josh Day', '[email protected]');
INSERT INTO employees (employee_id,employee_name, email_id) VALUES ('103','Sean Mann', '[email protected]');
INSERT INTO employees (employee_id,employee_name, email_id) VALUES ('104','Evan Blake', '[email protected]');
INSERT INTO employees (employee_id,employee_name, email_id) VALUES ('105','Toby Scott', '[email protected]');
SELECT * FROM employees;
SELECT email_id FROM
(
SELECT email_id, COUNT(email_id) AS nuOfAppearence
FROM employees
GROUP BY email_id
) AS countEmail
WHERE nuOfAppearence> 1;
Sample Output:
email_id | -------------| [email protected]|
Relational Algebra Expression:
Relational Algebra Tree:
Solution-1:
SELECT email_id
FROM employees
GROUP BY email_id
HAVING COUNT(email_id) > 1;
Solution-2:
SELECT DISTINCT p1.email_id
FROM employees p1, employees p2
WHERE p1.email_id = p2.email_id AND p1.employee_id != p2.employee_id;
SQL Code Editor:
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