Question

我用来做

SELECT email, COUNT(email) AS occurences
FROM wineries
GROUP BY email
HAVING (COUNT(email) > 1);

根据他们的电子邮件查找重复项。

但是现在我需要他们的ID来定义要删除哪一个。

第二个约束是：我只想要LAST INSERTED重复项。

因此，如果有2个条目将test@test.com作为电子邮件，并且它们的ID分别为40和12782，那么它将仅删除12782条目并保留40条条目。

关于我如何做到这一点的任何想法？我一直在混搭SQL大约一个小时，似乎无法找到确切的方法。

谢谢，祝你有愉快的一天！

Answer 1

好吧，你有点回答你的问题。您似乎想要max(id)：

SELECT email, COUNT(email) AS occurences, max(id)
FROM wineries
GROUP BY email
HAVING (COUNT(email) > 1);

您可以使用该语句删除其他人。使用join删除有一个棘手的语法，您必须首先列出表名，然后使用连接指定from子句：

delete wineries
            from wineries join
            (select email, max(id) as maxid
             from wineries
             group by email
             having count(*) > 1
            ) we
            on we.email = wineries.email and
               wineries.id < we.maxid;

或者将其写为exists子句：

delete from wineries
    where exists (select 1
                  from (select email, max(id) as maxid
                        from wineries
                        group by email
                       ) we
                  where we.email = wineries.email and wineries.id < we.maxid
                 )

Answer 2

select email, max(id), COUNT(email) AS occurences
FROM wineries
GROUP BY email
HAVING (COUNT(email) > 1);

Answer 3

delete from wineries
where id not in
(
  select * from 
  ( 
     select min(id)
     from wineries
     group by email
  ) x
)

你需要一个子查询来欺骗MySQL从它同时选择的表中删除。

Answer 4

DELETE duplicates.*
FROM wineries
JOIN wineries AS duplicates USING (email)
WHERE duplicates.id < wineries.id;

play with it on sqlfiddle.com

Answer 5

这是最简单的选项：

DELETE FROM wineries
 WHERE id NOT IN
(
  SELECT MIN(id) id
    FROM wineries
GROUP BY email 
);

这将仅保留每个电子邮件地址的第一个插入记录，所有其他记录将被删除。应该将这个答案归功于@juergen d，因为这只是他的答案的修订版。

使用MySQL查找最新的重复ID

5 个答案: