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

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。

其他回答

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

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;

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
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! !

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。