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

这将在一次表传递中选择重复项,没有子查询。

SELECT  *
FROM    (
        SELECT  ao.*, (@r := @r + 1) AS rn
        FROM    (
                SELECT  @_address := 'N'
                ) vars,
                (
                SELECT  *
                FROM
                        list a
                ORDER BY
                        address, id
                ) ao
        WHERE   CASE WHEN @_address <> address THEN @r := 0 ELSE 0 END IS NOT NULL
                AND (@_address := address ) IS NOT NULL
        ) aoo
WHERE   rn > 1

这个查询实际上模拟了Oracle和SQL Server中的ROW_NUMBER()

详见我博客上的文章:

分析函数:SUM, AVG, ROW_NUMBER -在MySQL模拟。

其他回答

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

为什么不直接INNER JOIN表本身呢?

SELECT a.firstname, a.lastname, a.address
FROM list a
INNER JOIN list b ON a.address = b.address
WHERE a.id <> b.id

如果地址可能存在两次以上,则需要DISTINCT。

Powerlord的答案确实是最好的,我建议再做一个改变:使用LIMIT来确保db不会超载:

SELECT firstname, lastname, list.address FROM list
INNER JOIN (SELECT address FROM list
GROUP BY address HAVING count(id) > 1) dup ON list.address = dup.address
LIMIT 10

如果没有WHERE和when连接,使用LIMIT是一个好习惯。从小值开始,检查查询有多重,然后增加限制。

    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 * FROM booking WHERE DATE(created_at) = '2022-01-11' 和代码在( 从预订中选择代码 按代码分组 have COUNT(code) > )由id DESC订购