方法是使用一个嵌套查询，每个重复查询一行，外部查询只返回内部查询结果的计数。

SELECT count(*) AS duplicate_count
FROM (
 SELECT name FROM tbl
 GROUP BY name HAVING COUNT(name) > 1
) AS t

使用IF语句获取所需的输出：

SELECT name, COUNT(*) AS times, IF (COUNT(*)>1,"duplicated", "not duplicated") AS duplicated FROM <MY_TABLE> GROUP BY name

输出：

AAA 2 duplicated
BBB 3 duplicated
CCC 1 not duplicated

对于列表：

SELECT COUNT(`name`) AS adet, name
FROM  `tbl` WHERE `status`=1 GROUP BY `name`
ORDER BY `adet`  DESC

总数量：

SELECT COUNT(*) AS Total
    FROM (SELECT COUNT(name) AS cou FROM tbl GROUP BY name HAVING cou>1 ) AS virtual_tbl

//共5条

为什么不把它 Package 在一个子查询中：

SELECT Count(*) TotalDups
FROM
(
    select Name, Count(*)
    from yourTable
    group by name
    having Count(*) > 1
) x

查看SQL Fiddle with Demo

接受的答案计算的是重复的行数，而不是重复的数量。如果要计算实际重复次数，请使用以下命令：

SELECT COALESCE(SUM(rows) - count(1), 0) as dupes FROM(

    SELECT COUNT(1) as rows
    FROM `yourtable`
    GROUP BY `name`
    HAVING rows > 1

) x

它的作用是将组中的重复项相加，然后减去具有重复项的记录的数量。原因是按总计分组并不都是重复的，这些分组中的每一个的一条记录是唯一的行。
小提琴：http：//sqlfiddle.com/#！2/29639a/3

SQL代码是：

SELECT VERSION_ID, PROJECT_ID, VERSION_NO, COUNT(VERSION_NO) AS dup_cnt
FROM MOVEMENTS
GROUP BY VERSION_NO
HAVING (dup_cnt > 1 && PROJECT_ID = 11660)

我用这个查询我自己的表在PHP中，但它只给我一个结果，而我想每个用户名重复的数量，这是可能的吗？

SELECT count(*) AS duplicate_count
FROM (
 SELECT username FROM login_history
 GROUP BY username HAVING COUNT(time) > 1
) AS t;

以下是查询：

select user_name, count(*) 
from users 
group by user_name 
having count(user_name) > 1

这将返回：
| 用户名|计数|
| - -----|- -----|
| ABC| 5个|
| xyz| 3|