Finding duplicate values in a SQL table

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

Simply group on both of the columns.

Note: the older ANSI standard is to have all non-aggregated columns in the GROUP BY but this has changed with the idea of "functional dependency":

In relational database theory, a functional dependency is a constraint between two sets of attributes in a relation from a database. In other words, functional dependency is a constraint that describes the relationship between attributes in a relation.

Support is not consistent:

• Recent PostgreSQL supports it.
• SQL Server (as at SQL Server 2017) still requires all non-aggregated columns in the GROUP BY.
• MySQL is unpredictable and you need sql_mode=only_full_group_by:
  • GROUP BY lname ORDER BY showing wrong results;
  • Which is the least expensive aggregate function in the absence of ANY() (see comments in accepted answer).
• Oracle isn't mainstream enough (warning: humour, I don't know about Oracle).

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)

Try this:

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

In contrast to other answers you can view the whole records containing all columns if there are any. In the PARTITION BY part of row_number function choose the desired unique/duplicit columns.

SELECT  *
FROM    (
 SELECT a.*
 ,      Row_Number() OVER (PARTITION BY Name, Age ORDER BY Name) AS r
 FROM   Customers AS a
)       AS b
WHERE   r > 1;

When you want to select ALL duplicated records with ALL fields you can write it like

CREATE TABLE test (
        id      bigint GENERATED ALWAYS AS IDENTITY PRIMARY KEY
,       c1      integer
,       c2      text
,       d       date DEFAULT now()
,       v       text
);

INSERT INTO test (c1, c2, v) VALUES
(1, 'a', 'Select'),
(1, 'a', 'ALL'),
(1, 'a', 'multiple'),
(1, 'a', 'records'),
(2, 'b', 'in columns'),
(2, 'b', 'c1 and c2'),
(3, 'c', '.');
SELECT * FROM test ORDER BY 1;

SELECT  *
FROM    test
WHERE   (c1, c2) IN (
 SELECT c1, c2
 FROM   test
 GROUP  BY 1,2
 HAVING count(*) > 1
)
ORDER   BY 1;

Tested in PostgreSQL.

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

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

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

This selects/deletes all duplicate records except one record from each group of duplicates. So, the delete leaves all unique records + one record from each group of the duplicates.

Select duplicates:

SELECT *
FROM table
WHERE
    id NOT IN (
        SELECT MIN(id)
        FROM table
        GROUP BY column1, column2
);

Delete duplicates:

DELETE FROM table
WHERE
    id NOT IN (
        SELECT MIN(id)
        FROM table
        GROUP BY column1, column2
);

Be aware of larger amounts of records, it can cause performance problems.

try this code

WITH CTE AS

( SELECT Id, Name, Age, Comments, RN = ROW_NUMBER()OVER(PARTITION BY Name,Age ORDER BY ccn)
FROM ccnmaster )
select * from CTE 

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', 'asd@asd.com');
insert into my_users values (2, 'Sam', 'asd@asd.com');
insert into my_users values (3, 'Tom', 'asd@asd.com');
insert into my_users values (4, 'Bob', 'bob@asd.com');
insert into my_users values (5, 'Tom', 'asd@asd.com');

commit;

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

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

If you wish to see if there is any duplicate rows in your table, I used below Query:

create table my_table(id int, name varchar(100), email varchar(100));

insert into my_table values (1, 'shekh', 'shekh@rms.com');
insert into my_table values (1, 'shekh', 'shekh@rms.com');
insert into my_table values (2, 'Aman', 'aman@rms.com');
insert into my_table values (3, 'Tom', 'tom@rms.com');
insert into my_table values (4, 'Raj', 'raj@rms.com');


Select COUNT(1) As Total_Rows from my_table 
Select Count(1) As Distinct_Rows from ( Select Distinct * from my_table) abc 

SELECT id, COUNT(id) FROM table1 GROUP BY id HAVING COUNT(id)>1;

I think this will work properly to search repeated values in a particular column.

select id,name,COUNT(*) from user group by Id,Name having COUNT(*)>1

 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
/

This is the easy thing I've come up with. It uses a common table expression (CTE) and a partition window (I think these features are in SQL 2008 and later).

This example finds all students with duplicate name and dob. The fields you want to check for duplication go in the OVER clause. You can include any other fields you want in the projection.

with cte (StudentId, Fname, LName, DOB, RowCnt)
as (
SELECT StudentId, FirstName, LastName, DateOfBirth as DOB, SUM(1) OVER (Partition By FirstName, LastName, DateOfBirth) as RowCnt
FROM tblStudent
)
SELECT * from CTE where RowCnt > 1
ORDER BY DOB, LName

How we can count the duplicated values?? either it is repeated 2 times or greater than 2. just count them, not group wise.

as simple as

select COUNT(distinct col_01) from Table_01

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

I think this will help you

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

Well this question has been answered very neatly in all the above answers. But I would like to list all the possible manners, we can do this in various ways which may impart the understanding how we can do it and seeker can pick one of the solution which best fits to his/her need as this is one of the most common query SQL developer come across different business usecases or sometime in interviews as well.

Creating Sample Data

I will start with setting up some sample data from this question only.

Create table NewTable (id int, name varchar(10), email varchar(50))
INSERT  NewTable VALUES (1,'John','asd@asd.com')
INSERT  NewTable VALUES (2,'Sam','asd@asd.com')
INSERT  NewTable VALUES (3,'Tom','asd@asd.com')
INSERT  NewTable VALUES (4,'Bob','bob@asd.com')
INSERT  NewTable VALUES (5,'Tom','asd@asd.com')

1. USING GROUP BY CLAUSE

SELECT
    name,email, COUNT(*) AS Occurence
    FROM NewTable
    GROUP BY name,email
    HAVING COUNT(*)>1

How it works:

• the GROUP BY clause groups the rows into groups by values in both name and email columns.
• Then, the COUNT() function returns the number of occurrences of each group (name,email).
• Then, the HAVING clause keeps only duplicate groups, which are groups that have more than one occurrence.

2. Using CTE:

To return the entire row for each duplicate row, you join the result of the above query with the NewTable table using a common table expression (CTE):

WITH cte AS (
    SELECT
        name, 
        email, 
        COUNT(*) occurrences
    FROM NewTable
    GROUP BY 
        name, 
        email
    HAVING COUNT(*) > 1
)
SELECT 
    t1.Id,
    t1.name, 
    t1.email
FROM  NewTable t1
    INNER JOIN cte ON 
        cte.name = t1.name AND 
        cte.email = t1.email
ORDER BY 
    t1.name, 
    t1.email;

3. Using ROW_NUMBER() function

WITH cte AS (
    SELECT 
        name, 
        email, 
        ROW_NUMBER() OVER (
            PARTITION BY name,email
            ORDER BY name,email) rownum
    FROM 
        NewTable t1
) 
SELECT 
  * 
FROM 
    cte 
WHERE 
    rownum > 1;

How it works:

• ROW_NUMBER() distributes rows of the NewTable table into partitions by values in the name and email columns. The duplicate rows will have repeated values in the name and email columns, but different row numbers
• Outer query removes the first row in each group.

Well Now I believe, you can have sound Idea of how to find duplicates and apply the logic to find duplicate in all possible scenarios. Thanks.

This should also work, maybe give it try.

  Select * from Users a
            where EXISTS (Select * from Users b 
                where (     a.name = b.name 
                        OR  a.email = b.email)
                     and a.ID != b.id)

Especially good in your case If you search for duplicates who have some kind of prefix or general change like e.g. new domain in mail. then you can use replace() at these columns

SELECT * FROM users u where rowid = (select max(rowid) from users u1 where
u.email=u1.email);

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

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;

If you want to find duplicate data (by one or several criterias) and select the actual rows.

with MYCTE as (
    SELECT DuplicateKey1
        ,DuplicateKey2 --optional
        ,count(*) X
    FROM MyTable
    group by DuplicateKey1, DuplicateKey2
    having count(*) > 1
) 
SELECT E.*
FROM MyTable E
JOIN MYCTE cte
ON E.DuplicateKey1=cte.DuplicateKey1
    AND E.DuplicateKey2=cte.DuplicateKey2
ORDER BY E.DuplicateKey1, E.DuplicateKey2, CreatedAt

http://developer.azurewebsites.net/2014/09/better-sql-group-by-find-duplicate-data/

To delete records whose names are duplicate

;WITH CTE AS    
(

    SELECT ROW_NUMBER() OVER (PARTITION BY name ORDER BY name) AS T FROM     @YourTable    
)

DELETE FROM CTE WHERE T > 1

To Check From duplicate Record in a table.

select * from users s 
where rowid < any 
(select rowid from users k where s.name = k.name and s.email = k.email);

or

select * from users s 
where rowid not in 
(select max(rowid) from users k where s.name = k.name and s.email = k.email);

To Delete the duplicate record in a table.

delete from users s 
where rowid < any 
(select rowid from users k where s.name = k.name and s.email = k.email);

or

delete from users s 
where rowid not in 
(select max(rowid) from users k where s.name = k.name and s.email = k.email);

Another easy way you can try this using analytic function as well:

SELECT * from 

(SELECT name, email,

COUNT(name) OVER (PARTITION BY name, email) cnt 

FROM users)

WHERE cnt >1;

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

You may want to try this

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