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
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;
其他回答
SELECT *
FROM (SELECT address, COUNT(id) AS cnt
FROM list
GROUP BY address
HAVING ( COUNT(id) > 1 ))
最快重复项删除查询过程:
/* create temp table with one primary column id */
INSERT INTO temp(id) SELECT MIN(id) FROM list GROUP BY (isbn) HAVING COUNT(*)>1;
DELETE FROM list WHERE id IN (SELECT id FROM temp);
DELETE FROM temp;
会是这样的:
SELECT t1.firstname t1.lastname t1.address FROM list t1
INNER JOIN list t2
WHERE
t1.id < t2.id AND
t1.address = t2.address;
就我个人而言，这个问题解决了我的问题:
SELECT `SUB_ID`, COUNT(SRV_KW_ID) as subscriptions FROM `SUB_SUBSCR` group by SUB_ID, SRV_KW_ID HAVING subscriptions > 1;
这个脚本所做的是在表中显示所有存在过一次以上的订阅者ID，以及找到的重复的数量。
这是表的列:
| SUB_SUBSCR_ID | int(11) | NO | PRI | NULL | auto_increment |
| MSI_ALIAS | varchar(64) | YES | UNI | NULL | |
| SUB_ID | int(11) | NO | MUL | NULL | |
| SRV_KW_ID | int(11) | NO | MUL | NULL | |
希望对你也有帮助!
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! !
