我想在MySQL数据库中取出重复的记录。这可以用:

SELECT address, count(id) as cnt FROM list
GROUP BY address HAVING cnt > 1

结果是:

100 MAIN ST    2

我想要拖动它,以便它显示复制的每一行。喜欢的东西:

JIM    JONES    100 MAIN ST
JOHN   SMITH    100 MAIN ST

有什么想法可以做到吗?我试图避免做第一个,然后在代码中用第二个查询查找重复。


当前回答

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。)

其他回答

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。

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 * 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

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