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 `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! !
其他回答
SELECT firstname, lastname, address FROM list
WHERE
Address in
(SELECT address FROM list
GROUP BY address
HAVING count(*) > 1)
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;
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;
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 `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! !
