在MySQL中查找重复的记录

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 * 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。

其他回答

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

 SELECT firstname, lastname, address FROM list
 WHERE 
 Address in 
 (SELECT address FROM list
 GROUP BY address
 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

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”)。

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