SELECT username, email, COUNT(*)
FROM users
GROUP BY username, email
HAVING COUNT(*) > 1
SELECT name, COUNT(email)
FROM users
GROUP BY email
HAVING COUNT(email) > 1
CREATE TABLE new_table LIKE original_table;
INSERT INTO new_table SELECT * FROM original_table;
Multiple field=
SELECT username, email, COUNT(*)
FROM users
GROUP BY username, email
HAVING COUNT(*) > 1
Single field=
SELECT _column, COUNT(*)
FROM _table
GROUP BY _column
HAVING COUNT(*) > 1
• SELECT first_name, COUNT (first_name) FROM employees
GROUP BY first_name
HAVING (COUNT(first_name) > 1);
SELECT column1, column2, COUNT(*) as 'Count'
FROM tablename
GROUP BY column1, column2
HAVING Count(*) > 1
ORDER BY [Count] DESC;
/* This SQL Query will show each of those duplicate
rows individually instead of just grouping it */
SELECT username,email
FROM `users`
WHERE `username`
IN (SELECT username FROM `users` GROUP BY username HAVING COUNT(username) > 1)
SELECT DISTINCT CITY FROM STATION WHERE ID%2=0;
/*SELECT hotel_id, reservation_id, COUNT(hotel_id)FROM staywhere reservation_id is not nullGROUP BY hotel_id, reservation_idHAVING COUNT(hotel_id) > 1*/select distinct s.hotel_id , t.* from stay sright join ( select hotel_id, reservation_id, count(*) as qty from staywhere reservation_id !=1 group by hotel_id, reservation_id having count(*) > 1) t on s.hotel_id = t.hotel_id and s.reservation_id = t.reservation_id