如何将SQL结果集限制为不太常见的项目

Question

问题：我有一个名称和地址列表。一些姓名（人）与其他人一样具有相同的地址（街道，邮政编码，城镇）。我想选择地址不超过三次的所有这些名称，其余的前三个名称分别指向同一个地址。 例子：

 
Albert | Adr1 
Berta | Adr1 
Cesar | Adr1 
Donald | Adr1 
Eric | Adr2 
Fritz | Adr2 
Gerd | Adr2 
Henry | Adr3

结果集应该是

Albert | Adr1 
Berta | Adr1 
Cesar | Adr1 
Eric | Adr2 
Fritz | Adr2 
Gerd | Adr2 
Henry | Adr3

唐纳德丢失，因为他是用同一个地址组的第4位。 这个结果可以通过UNION和子查询来实现吗？像

select * from addresses where address in 
(select address from addresses group by address having count(address) <= 3) 
UNION 
select * from addresses where address in 
(select address from addresses group by address having count(address) > 3 limit 3)

事我知道，这个查询是错误的，因为它限制了整个结果集的地址超过3个出现次数。 我不知道这是否可以用UNION和子查询在单个SELECT中完成。我现在将用PHP/MySQL程序执行它，但只是为了好玩，会对仅限SQL的解决方案感兴趣。

我看了一下SQL query with limit on rows from one table, not the result set，但这并不反映我的情况 - 或者不是吗？

Answer 1

你可以尝试像

SELECT PersonName, 
     Address 
FROM (
      SELECT *, 
        (SELECT COUNT(1) FROM addresses WHERE Address = a.Address AND PersonName < a.PersonName) CountLess 
      FROM addresses a 
     ) sub 
WHERE sub.CountLess <= 2 

Answer 2

select name, address 
from 
(select *, row_number() over (partition by address order by name) as namenum 
    from yourTable 
) t 
where namenum <= 3;