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

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

如果你想删除重复使用DISTINCT

否则使用这个查询:

SELECT users.*，COUNT(user_ID) as user FROM users GROUP BY user_name有用户>

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

用你的表格替换城市。 将name替换为字段名

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;