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

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

要查找Employee中的name列中有多少记录是重复的，下面的查询很有用;

Select name from employee group by name having count(*)>1;

我更喜欢使用窗口函数(MySQL 8.0+)来查找副本，因为我可以看到整行:

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小提琴演示

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