Finding duplicate values in a SQL table

Question

It\'s easy to find duplicates with one field:

SELECT name, COUNT(email) 
FROM users
GROUP BY email
HAVING COUNT(email) > 1

So if we have

Answer 1

If you want to delete the duplicates, here's a much simpler way to do it than having to find even/odd rows into a triple sub-select:

SELECT id, name, email 
FROM users u, users u2
WHERE u.name = u2.name AND u.email = u2.email AND u.id > u2.id

And so to delete:

DELETE FROM users
WHERE id IN (
    SELECT id/*, name, email*/
    FROM users u, users u2
    WHERE u.name = u2.name AND u.email = u2.email AND u.id > u2.id
)

Much more easier to read and understand IMHO

Note: The only issue is that you have to execute the request until there is no rows deleted, since you delete only 1 of each duplicate each time

Answer 2

SELECT name, email,COUNT(email) 
FROM users 
WHERE email IN (
    SELECT email 
    FROM users 
    GROUP BY email 
    HAVING COUNT(email) > 1)

Answer 3

SELECT column_name,COUNT(*) FROM TABLE_NAME GROUP BY column1, HAVING COUNT(*) > 1;

Answer 4

 select emp.ename, emp.empno, dept.loc 
          from emp
 inner join dept 
          on dept.deptno=emp.deptno
 inner join
    (select ename, count(*) from
    emp
    group by ename, deptno
    having count(*) > 1)
 t on emp.ename=t.ename order by emp.ename
/

Answer 5

You may want to try this

SELECT NAME, EMAIL, COUNT(*)
FROM USERS
GROUP BY 1,2
HAVING COUNT(*) > 1

Answer 6

The most important thing here is to have the fastest function. Also indices of duplicates should be identified. Self join is a good option but to have a faster function it is better to first find rows that have duplicates and then join with original table for finding id of duplicated rows. Finally order by any column except id to have duplicated rows near each other.

SELECT u.*
FROM users AS u
JOIN (SELECT username, email
      FROM users
      GROUP BY username, email
      HAVING COUNT(*)>1) AS w
ON u.username=w.username AND u.email=w.email
ORDER BY u.email;