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 * 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。

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!

    Find duplicate Records:

    Suppose we have table : Student 
    student_id int
    student_name varchar
    Records:
    +------------+---------------------+
    | student_id | student_name        |
    +------------+---------------------+
    |        101 | usman               |
    |        101 | usman               |
    |        101 | usman               |
    |        102 | usmanyaqoob         |
    |        103 | muhammadusmanyaqoob |
    |        103 | muhammadusmanyaqoob |
    +------------+---------------------+

    Now we want to see duplicate records
    Use this query:


   select student_name,student_id ,count(*) c from student group by student_id,student_name having c>1;

+--------------------+------------+---+
| student_name        | student_id | c |
+---------------------+------------+---+
| usman               |        101 | 3 |
| muhammadusmanyaqoob |        103 | 2 |
+---------------------+------------+---+
SELECT t.*,(select count(*) from city as tt where tt.name=t.name) as count FROM `city` as t where (select count(*) from city as tt where tt.name=t.name) > 1 order by count desc

用你的表格替换城市。 将name替换为字段名

SELECT date FROM logs group by date having count(*) >= 2