Question

好朋友，我有这个查询有效，但执行时间很长。我想知道是否有更快的方法来实现这一点。

SELECT id, email FROM member WHERE email IN 
(SELECT email FROM member GROUP BY email HAVING count( * ) >1 ) 
ORDER BY `email` ASC

基本上，有些条目中同一封电子邮件不止一次出现，我只是希望在“电子邮件”重复的条目中返回这些行。

上述查询在这个方向上有效，但是很长。

干杯。

Answer 1

您可以先将结果分组，然后将它们连接到成员表，以确保只显示包含重复电子邮件的行。

SELECT m.id, m.email
FROM member m JOIN (
    SELECT email 
    FROM member 
    GROUP BY email 
    HAVING COUNT(*) > 1
  ) g ON m.email = g.email
ORDER BY m.email ASC

Answer 2

由于嵌套选择，您的查询很慢，每次都会重新计算。最好的解决方案是重写您的算法，以便您可以使用如下查询：

SELECT id, email 
FROM member GROUP BY email
HAVING count( * ) >1
ORDER BY `email`

不幸的是，你回来的id将是每个组中的随机选择。这可能是一个更有用的查询：

SELECT GROUP_CONCAT(id), email 
FROM member GROUP BY email
HAVING count( * ) >1
ORDER BY `email`

Answer 3

你能分两个阶段做到吗？首先创建一个包含＆gt;所有电子邮件的临时表1出现，然后通过电子邮件字段将主表连接到临时表...

如果您的成员表在电子邮件字段中有索引，那么这应该非常快。


CREATE TEMPORARY TABLE ddd
SELECT email, count(*) as cnt FROM member GROUP BY email HAVING cnt>1;

SELECT * FROM ddd
INNER JOIN member USING (email);

Answer 4

当您只需要做一次

时，您正在进行两次查询

SELECT id, email 
FROM member
GROUP BY email 
HAVING count( * ) > 1
ORDER BY `email` ASC

Answer 5

select id,email,count(*) as n from member group by id having n > 1;

帮助优化这个mysql查询

5 个答案: