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

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;

其他回答

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;
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。

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。)

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。

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。