Finding duplicate values in MySQL

Do a SELECT with a GROUP BY clause. Let's say name is the column you want to find duplicates in:

SELECT name, COUNT(*) c FROM table GROUP BY name HAVING c > 1;

This will return a result with the name value in the first column, and a count of how many times that value appears in the second.

SELECT varchar_col
FROM table
GROUP BY varchar_col
HAVING COUNT(*) > 1;

SELECT  *
FROM    mytable mto
WHERE   EXISTS
        (
        SELECT  1
        FROM    mytable mti
        WHERE   mti.varchar_column = mto.varchar_column
        LIMIT 1, 1
        )

This query returns complete records, not just distinct varchar_column's.

This query doesn't use COUNT(*). If there are lots of duplicates, COUNT(*) is expensive, and you don't need the whole COUNT(*), you just need to know if there are two rows with same value.

This is achieved by the LIMIT 1, 1 at the bottom of the correlated query (essentially meaning "return the second row"). EXISTS would only return true if the aforementioned second row exists (i. e. there are at least two rows with the same value of varchar_column) .

Having an index on varchar_column will, of course, speed up this query greatly.

to get all the data that contains duplication i used this:

SELECT * FROM TableName INNER JOIN(
  SELECT DupliactedData FROM TableName GROUP BY DupliactedData HAVING COUNT(DupliactedData) > 1 order by DupliactedData)
  temp ON TableName.DupliactedData = temp.DupliactedData;

TableName = the table you are working with.

DupliactedData = the duplicated data you are looking for.

Assuming your table is named TableABC and the column which you want is Col and the primary key to T1 is Key.

SELECT a.Key, b.Key, a.Col 
FROM TableABC a, TableABC b
WHERE a.Col = b.Col 
AND a.Key <> b.Key

The advantage of this approach over the above answer is it gives the Key.

I am not seeing any JOIN approaches, which have many uses in terms of duplicates.

This approach gives you actual doubled results.

SELECT t1.* FROM my_table as t1 
LEFT JOIN my_table as t2 
ON t1.name=t2.name and t1.id!=t2.id 
WHERE t2.id IS NOT NULL 
ORDER BY t1.name

I prefer to use windowed functions(MySQL 8.0+) to find duplicates because I could see entire row:

WITH cte AS (
  SELECT *
    ,COUNT(*) OVER(PARTITION BY col_name) AS num_of_duplicates_group
    ,ROW_NUMBER() OVER(PARTITION BY col_name ORDER BY col_name2) AS pos_in_group
  FROM table
)
SELECT *
FROM cte
WHERE num_of_duplicates_group > 1;

DB Fiddle Demo

SELECT ColumnA, COUNT( * )
FROM Table
GROUP BY ColumnA
HAVING COUNT( * ) > 1

I improved from this:

SELECT 
    col, 
    COUNT(col)
FROM
    table_name
GROUP BY col
HAVING COUNT(col) > 1; 

As a variation on Levik's answer that allows you to find also the ids of the duplicate results, I used the following:

SELECT * FROM table1 WHERE column1 IN (SELECT column1 AS duplicate_value FROM table1 GROUP BY column1 HAVING COUNT(*) > 1)

If you want to remove duplicate use DISTINCT

Otherwise use this query:

SELECT users.*,COUNT(user_ID) as user FROM users GROUP BY user_name HAVING user > 1;

Thanks to @novocaine for his great answer and his solution worked for me. I altered it slightly to include a percentage of the recurring values, which was needed in my case. Below is the altered version. It reduces the percentage to two decimal places. If you change the ,2 to 0, it will display no decimals, and to 1, then it will display one decimal place, and so on.

SELECT GROUP_CONCAT(id), name, COUNT(*) c, 
COUNT(*) OVER() AS totalRecords, 
CONCAT(FORMAT(COUNT(*)/COUNT(*) OVER()*100,2),'%') as recurringPecentage
FROM table
GROUP BY name
HAVING c > 1

Try using this query:

SELECT name, COUNT(*) value_count FROM company_master GROUP BY name HAVING value_count > 1;