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 a table

ID   NAME   EMAIL
1    John   [e-mail]
2    Sam    [e-mail]
3    Tom    [e-mail]
4    Bob    [e-mail]
5    Tom    [e-mail]

This query will give us John, Sam, Tom, Tom because they all have the same email.

However, what I want is to get duplicates with the same email and name.

That is, I want to get "Tom", "Tom".

The reason I need this: I made a mistake, and allowed to insert duplicate name and email values. Now I need to remove/change the duplicates, so I need to find them first.

Answer 1

select name, email
, case 
when ROW_NUMBER () over (partition by name, email order by name) > 1 then 'Yes'
else 'No'
end "duplicated ?"
from users

Answer 2

By Using CTE also we can find duplicate value like this

with MyCTE
as
(
select Name,EmailId,ROW_NUMBER() over(PARTITION BY EmailId order by id) as Duplicate from [Employees]

)
select * from MyCTE where Duplicate>1

Answer 3

try this:

declare @YourTable table (id int, name varchar(10), email varchar(50))

INSERT @YourTable VALUES (1,'John','John-email')
INSERT @YourTable VALUES (2,'John','John-email')
INSERT @YourTable VALUES (3,'fred','John-email')
INSERT @YourTable VALUES (4,'fred','fred-email')
INSERT @YourTable VALUES (5,'sam','sam-email')
INSERT @YourTable VALUES (6,'sam','sam-email')

SELECT
    name,email, COUNT(*) AS CountOf
    FROM @YourTable
    GROUP BY name,email
    HAVING COUNT(*)>1

OUTPUT:

name       email       CountOf
---------- ----------- -----------
John       John-email  2
sam        sam-email   2

(2 row(s) affected)

if you want the IDs of the dups use this:

SELECT
    y.id,y.name,y.email
    FROM @YourTable y
        INNER JOIN (SELECT
                        name,email, COUNT(*) AS CountOf
                        FROM @YourTable
                        GROUP BY name,email
                        HAVING COUNT(*)>1
                    ) dt ON y.name=dt.name AND y.email=dt.email

OUTPUT:

id          name       email
----------- ---------- ------------
1           John       John-email
2           John       John-email
5           sam        sam-email
6           sam        sam-email

(4 row(s) affected)

to delete the duplicates try:

DELETE d
    FROM @YourTable d
        INNER JOIN (SELECT
                        y.id,y.name,y.email,ROW_NUMBER() OVER(PARTITION BY y.name,y.email ORDER BY y.name,y.email,y.id) AS RowRank
                        FROM @YourTable y
                            INNER JOIN (SELECT
                                            name,email, COUNT(*) AS CountOf
                                            FROM @YourTable
                                            GROUP BY name,email
                                            HAVING COUNT(*)>1
                                        ) dt ON y.name=dt.name AND y.email=dt.email
                   ) dt2 ON d.id=dt2.id
        WHERE dt2.RowRank!=1
SELECT * FROM @YourTable

OUTPUT:

id          name       email
----------- ---------- --------------
1           John       John-email
3           fred       John-email
4           fred       fred-email
5           sam        sam-email

(4 row(s) affected)

Answer 4

Try this:

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

Answer 5

In case you work with Oracle, this way would be preferable:

create table my_users(id number, name varchar2(100), email varchar2(100));

insert into my_users values (1, 'John', '[e-mail]');
insert into my_users values (2, 'Sam', '[e-mail]');
insert into my_users values (3, 'Tom', '[e-mail]');
insert into my_users values (4, 'Bob', '[e-mail]');
insert into my_users values (5, 'Tom', '[e-mail]');

commit;

select *
  from my_users
 where rowid not in (select min(rowid) from my_users group by name, email);

Answer 6

A little late to the party but I found a really cool workaround to finding all duplicate IDs:

SELECT GROUP_CONCAT( id )
FROM users
GROUP BY email
HAVING ( COUNT(email) > 1 )