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

    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 |
+---------------------+------------+---+

其他回答

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

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;
    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 |
+---------------------+------------+---+

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;