想象你的表 test 包含以下数据:

Imagine your table test contains the following data:

  select id, email
    from test;

ID                     EMAIL                
---------------------- -------------------- 
1                      aaa                  
2                      bbb                  
3                      ccc                  
4                      bbb                  
5                      ddd                  
6                      eee                  
7                      aaa                  
8                      aaa                  
9                      eee 

所以，我们需要找到所有重复的邮件并删除所有邮件，但要删除最新的 id.
在这种情况下，aaa、bbb 和eee 是重复的，所以我们要删除 ID 1、7、2 和 6.

So, we need to find all repeated emails and delete all of them, but the latest id.
In this case, aaa, bbb and eee are repeated, so we want to delete IDs 1, 7, 2 and 6.

要做到这一点，首先我们需要找到所有重复的电子邮件:

To accomplish this, first we need to find all the repeated emails:

      select email 
        from test
       group by email
      having count(*) > 1;

EMAIL                
-------------------- 
aaa                  
bbb                  
eee  

然后，从这个数据集中，我们需要为这些重复的电子邮件中的每一个找到最新的 id:

Then, from this dataset, we need to find the latest id for each one of these repeated emails:

  select max(id) as lastId, email
    from test
   where email in (
              select email 
                from test
               group by email
              having count(*) > 1
       )
   group by email;

LASTID                 EMAIL                
---------------------- -------------------- 
8                      aaa                  
4                      bbb                  
9                      eee                                 

最后，我们现在可以删除所有这些 Id 小于 LASTID 的电子邮件.所以解决方案是:

Finally we can now delete all of these emails with an Id smaller than LASTID. So the solution is:

delete test
  from test
 inner join (
  select max(id) as lastId, email
    from test
   where email in (
              select email 
                from test
               group by email
              having count(*) > 1
       )
   group by email
) duplic on duplic.email = test.email
 where test.id < duplic.lastId;

我现在没有在这台机器上安装 mySql，但应该可以工作

I don't have mySql installed on this machine right now, but should work

上面的删除有效，但我发现了一个更优化的版本:

The above delete works, but I found a more optimized version:

 delete test
   from test
  inner join (
     select max(id) as lastId, email
       from test
      group by email
     having count(*) > 1) duplic on duplic.email = test.email
  where test.id < duplic.lastId;

您可以看到它删除了最旧的重复项，即 1、7、2、6:

You can see that it deletes the oldest duplicates, i.e. 1, 7, 2, 6:

select * from test;
+----+-------+
| id | email |
+----+-------+
|  3 | ccc   |
|  4 | bbb   |
|  5 | ddd   |
|  8 | aaa   |
|  9 | eee   |
+----+-------+

另一个版本，是由 Rene Limon

delete from test
 where id not in (
    select max(id)
      from test
     group by email)