感谢@novocaine的精彩回答，他的解决方案对我很有效。我稍微改变了它，以包括一个百分比的循环值，这在我的例子中是需要的。以下是修改后的版本。它将百分比减少到小数点后两位。如果你把2改成0，它就不会显示小数，改成1，它就会显示一位小数，以此类推。

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

下面的代码将找到所有使用了不止一次的product_id。每个product_id只能得到一条记录。

SELECT product_id FROM oc_product_reward GROUP BY product_id HAVING count( product_id ) >1

代码取自:http://chandreshrana.blogspot.in/2014/12/find-duplicate-records-based-on-any.html

SELECT * 
FROM `dps` 
WHERE pid IN (SELECT pid FROM `dps` GROUP BY pid HAVING COUNT(pid)>1)

作为利维克的答案的一个变体，它可以让你找到重复结果的id，我使用了以下方法:

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

进一步采取@maxyfc的答案，我需要找到所有返回的重复值的行，这样我就可以在MySQL Workbench中编辑它们:

SELECT * FROM table
   WHERE field IN (
     SELECT field FROM table GROUP BY field HAVING count(*) > 1
   ) ORDER BY field

SELECT 
    t.*,
    (SELECT COUNT(*) FROM city AS tt WHERE tt.name=t.name) AS count 
FROM `city` AS t 
WHERE 
    (SELECT count(*) FROM city AS tt WHERE tt.name=t.name) > 1 ORDER BY count DESC