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

其他回答

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;

这样不是更简单吗?

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

?

    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;

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。