Question

我正在尝试获取所有我的帖子，其中投票IPAddress与我自己的匹配。这是我的SQL。问题是，当我和其他人投票赞成时，我得到了投票。我认为这段代码说左边连接是可能的选择帖子，而且没有IPAddr = =我的。但这似乎并没有发生。我怎么写这个？

select * from Post 
left join Vote as V on V.post=Post.id AND V.IPAddr<>123
where flag='1';

这是一些虚拟数据来说明问题。

create table P(id int primary key, body text);
create table V(id int primary key, val int, ip int, post int);
insert into P values(1,"aaa");
select * from P left join V on V.post=P.id where (V.ip is null or V.ip<>123);
insert into V values(1, 2,123,1);
select * from P left join V on V.post=P.id where (V.ip is null or V.ip<>123);
insert into V values(2, 2,13,1);
select * from P left join V on V.post=P.id where (V.ip is null or V.ip<>123);

Answer 1

select * 
from Post p 
left join Vote as V on V.post=p.id 
where (V.IPAddr is null or V.IPAddr<>123) 
    and flag='1';

<强>更新

select * 
from P 
left outer join V on V.post=P.id 
where not exists (
    select 1 from v
    where IP = 123
    and post = p.id
)

Answer 2

这是你需要的吗？

SELECT * FROM P 
WHERE NOT EXISTS
     (SELECT * FROM V 
      WHERE V.post=P.id AND V.ip=123)

Answer 3

也许就像......

SELECT *
    FROM P
    WHERE EXISTS (SELECT *
                      FROM V
                      WHERE V.IP IS NOT NULL AND
                            V.IP <> 123      AND
                            P.ID = V.POST);

正是您要找的。

由于左连接导致的结果不佳

3 个答案: