不是很有效，但应该可以工作:

SELECT *
FROM list AS outer
WHERE (SELECT COUNT(*)
        FROM list AS inner
        WHERE inner.address = outer.address) > 1;

关键是重写这个查询，以便它可以用作子查询。

SELECT firstname, 
   lastname, 
   list.address 
FROM list
   INNER JOIN (SELECT address
               FROM   list
               GROUP  BY address
               HAVING COUNT(id) > 1) dup
           ON list.address = dup.address;

这将在一次表传递中选择重复项，没有子查询。

SELECT  *
FROM    (
        SELECT  ao.*, (@r := @r + 1) AS rn
        FROM    (
                SELECT  @_address := 'N'
                ) vars,
                (
                SELECT  *
                FROM
                        list a
                ORDER BY
                        address, id
                ) ao
        WHERE   CASE WHEN @_address <> address THEN @r := 0 ELSE 0 END IS NOT NULL
                AND (@_address := address ) IS NOT NULL
        ) aoo
WHERE   rn > 1

这个查询实际上模拟了Oracle和SQL Server中的ROW_NUMBER()

详见我博客上的文章:

分析函数:SUM, AVG, ROW_NUMBER -在MySQL模拟。

    SELECT *
    FROM (SELECT  address, COUNT(id) AS cnt
    FROM list
    GROUP BY address
    HAVING ( COUNT(id) > 1 ))

为什么不直接INNER JOIN表本身呢?

SELECT a.firstname, a.lastname, a.address
FROM list a
INNER JOIN list b ON a.address = b.address
WHERE a.id <> b.id

如果地址可能存在两次以上，则需要DISTINCT。

 SELECT firstname, lastname, address FROM list
 WHERE 
 Address in 
 (SELECT address FROM list
 GROUP BY address
 HAVING count(*) > 1)

select `cityname` from `codcities` group by `cityname` having count(*)>=2

这是你问的类似的问题，它是200%的工作和简单。 享受! !

寻找重复地址比看起来要复杂得多，特别是当你要求准确性时。在这种情况下，一个MySQL查询是不够的…

我在SmartyStreets工作，在那里我们解决验证和重复数据删除等问题，我看到过很多类似问题的不同挑战。

有一些第三方服务会在列表中为你标记重复项。仅使用MySQL子查询来执行此操作不会考虑地址格式和标准的差异。美国邮政总局(USPS)有一定的指导方针来制定这些标准，但只有少数供应商获得了执行此类操作的认证。

因此，我建议您最好的答案是，例如，将表导出到CSV文件中，并将其提交给有能力的列表处理程序。其中一个是LiveAddress，它会在几秒钟到几分钟内自动为你完成。它将用一个名为“duplicate”的新字段和一个Y值标记重复的行。

通过此查询通过电子邮件地址查找重复用户…

SELECT users.name, users.uid, users.mail, from_unixtime(created)
FROM users
INNER JOIN (
  SELECT mail
  FROM users
  GROUP BY mail
  HAVING count(mail) > 1
) dupes ON users.mail = dupes.mail
ORDER BY users.mail;

SELECT date FROM logs group by date having count(*) >= 2

另一个解决方案是使用表别名，如下所示:

SELECT p1.id, p2.id, p1.address
FROM list AS p1, list AS p2
WHERE p1.address = p2.address
AND p1.id != p2.id

在这种情况下，您真正要做的是获取原始的列表表，从中创建两个假装的表——p1和p2，然后在地址列上执行连接(第3行)。第4行确保相同的记录不会在结果集中多次出现(“重复重复”)。

select * from table_name t1 inner join (select distinct <attribute list> from table_name as temp)t2 where t1.attribute_name = t2.attribute_name

对于你的桌子来说，应该是这样的

select * from list l1 inner join (select distinct address from list as list2)l2 where l1.address=l2.address

这个查询将为您提供列表表中所有不同的地址条目…我不知道这将如何工作，如果你有任何主键值的名称等。

最快重复项删除查询过程:

/* create temp table with one primary column id */
INSERT INTO temp(id) SELECT MIN(id) FROM list GROUP BY (isbn) HAVING COUNT(*)>1;
DELETE FROM list WHERE id IN (SELECT id FROM temp);
DELETE FROM temp;

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

我们也可以发现重复依赖于不止一个字段。对于这些情况，您可以使用下面的格式。

SELECT COUNT(*), column1, column2 
FROM tablename
GROUP BY column1, column2
HAVING COUNT(*)>1;

就我个人而言，这个问题解决了我的问题:

SELECT `SUB_ID`, COUNT(SRV_KW_ID) as subscriptions FROM `SUB_SUBSCR` group by SUB_ID, SRV_KW_ID HAVING subscriptions > 1;

这个脚本所做的是在表中显示所有存在过一次以上的订阅者ID，以及找到的重复的数量。

这是表的列:

| SUB_SUBSCR_ID | int(11)     | NO   | PRI | NULL    | auto_increment |
| MSI_ALIAS     | varchar(64) | YES  | UNI | NULL    |                |
| SUB_ID        | int(11)     | NO   | MUL | NULL    |                |    
| SRV_KW_ID     | int(11)     | NO   | MUL | NULL    |                |

希望对你也有帮助!

我试着用最好的答案来回答这个问题，但它还是把我弄糊涂了。实际上，我只需要在表的一个字段上。下面这个链接的例子对我来说效果很好:

SELECT COUNT(*) c,title FROM `data` GROUP BY title HAVING c > 1;

从列表中选择地址where address = any (Select address from (Select address, count(id) CNT from list group by address with CNT > 1) as t1)按地址排序

内部子查询返回具有重复地址的行 外层子查询返回重复地址的地址列。 外层子查询必须只返回一列，因为它被用作操作符'= any'的操作数。

Powerlord的答案确实是最好的，我建议再做一个改变:使用LIMIT来确保db不会超载:

SELECT firstname, lastname, list.address FROM list
INNER JOIN (SELECT address FROM list
GROUP BY address HAVING count(id) > 1) dup ON list.address = dup.address
LIMIT 10

如果没有WHERE和when连接，使用LIMIT是一个好习惯。从小值开始，检查查询有多重，然后增加限制。

这还将显示有多少重复项，并将在没有连接的情况下对结果进行排序

SELECT  `Language` , id, COUNT( id ) AS how_many
FROM  `languages` 
GROUP BY  `Language` 
HAVING how_many >=2
ORDER BY how_many DESC

    Find duplicate Records:

    Suppose we have table : Student 
    student_id int
    student_name varchar
    Records:
    +------------+---------------------+
    | student_id | student_name        |
    +------------+---------------------+
    |        101 | usman               |
    |        101 | usman               |
    |        101 | usman               |
    |        102 | usmanyaqoob         |
    |        103 | muhammadusmanyaqoob |
    |        103 | muhammadusmanyaqoob |
    +------------+---------------------+

    Now we want to see duplicate records
    Use this query:


   select student_name,student_id ,count(*) c from student group by student_id,student_name having c>1;

+--------------------+------------+---+
| student_name        | student_id | c |
+---------------------+------------+---+
| usman               |        101 | 3 |
| muhammadusmanyaqoob |        103 | 2 |
+---------------------+------------+---+

这样不是更简单吗?

SELECT *
FROM tc_tariff_groups
GROUP BY group_id
HAVING COUNT(group_id) >1

?

要快速查看重复的行，可以运行一个简单的查询

在这里，我正在查询表，并列出所有重复的行相同的user_id, market_place和sku:

select user_id, market_place,sku, count(id)as totals from sku_analytics group by user_id, market_place,sku having count(id)>1;

要删除重复的行，必须决定要删除哪一行。例如id较低(通常较旧)或其他日期信息。在我的情况下，我只是想删除较低的id，因为较新的id是最新的信息。

首先仔细检查是否正确的记录将被删除。在这里，我正在选择将被删除的副本中的记录(通过唯一id)。

select a.user_id, a.market_place,a.sku from sku_analytics a inner join sku_analytics b where a.id< b.id and a.user_id= b.user_id and a.market_place= b.market_place and a.sku = b.sku;

然后我运行delete查询来删除dupes:

delete a from sku_analytics a inner join sku_analytics b where a.id< b.id and a.user_id= b.user_id and a.market_place= b.market_place and a.sku = b.sku;

备份，双重检查，验证，验证备份，然后执行。

我使用以下方法:

SELECT * FROM mytable
WHERE id IN (
  SELECT id FROM mytable
  GROUP BY column1, column2, column3
  HAVING count(*) > 1
)

SELECT * FROM booking WHERE DATE(created_at) = '2022-01-11' 和代码在( 从预订中选择代码 按代码分组 have COUNT(code) > )由id DESC订购

这里的大多数答案不适用于有多个重复结果和/或有多个列要检查重复的情况。当你在这种情况下，你可以使用这个查询来获得所有重复的id:

SELECT address, email, COUNT(*) AS QUANTITY_DUPLICATES, GROUP_CONCAT(id) AS ID_DUPLICATES
    FROM list
    GROUP BY address, email
    HAVING COUNT(*)>1;

如果希望将每个结果作为一行列出，则需要更复杂的查询。这是我发现有效的方法:

CREATE TEMPORARY TABLE IF NOT EXISTS temptable AS (    
    SELECT GROUP_CONCAT(id) AS ID_DUPLICATES
    FROM list
    GROUP BY address, email
    HAVING COUNT(*)>1
); 
SELECT d.* 
    FROM list AS d, temptable AS t 
    WHERE FIND_IN_SET(d.id, t.ID_DUPLICATES) 
    ORDER BY d.id;

SELECT id, count(*) as c  
 FROM 'list'
GROUP BY id HAVING c > 1

这将返回id和该id重复的次数，或者什么都没有，在这种情况下，您将不会有重复的id。

通过(例如:address)更改组中的id，它将返回一个地址由第一个找到的id与该地址重复标识的次数。

SELECT id, count(*) as c  
 FROM 'list'
GROUP BY address HAVING c > 1

我希望这能有所帮助。喜欢。)

会是这样的:

SELECT  t1.firstname t1.lastname t1.address FROM list  t1
    INNER JOIN  list t2 
    WHERE 
        t1.id < t2.id AND 
        t1.address = t2.address;