Question

我有3个表：用户，社区＆amp;讯息。

我想检索Joe不关注(id_user=1)

的用户帖子

用户

    id_user  |  name
       1        Joe
       2        Doe
       3        Moe
       4        Roe
       5        Clin

社区

  id_follower | id_followed
       1            3
       1            5

帖子

    id_post   | id_user | post  
      24            4       hi
      25            5       hello
      26            1       how are you
      27            3       come on
      28            4       let go
      29            2       get out

我期待检索的是

      24            4       hi - by Roe
      28            4       let go  - by Roe
      29            2       get out  - by Doe

我试过这个但没有工作

SELECT p*
FROM community as c
LEFT JOIN users as u ON u.id_user=c.id_followed
LEFT JOIN posts as p ON p.id_user!=c.id_followed 
WHERE c.id_follower=1 AND u.id_user!=1

Answer 1

这应该有效：

select * from Posts 
where id_user not in
(select id_followed from Community where id_follower = 1)

Answer 2

没有必要给出答案，让我们逐步完成逻辑。还有其他方法可以做到这一点，但是因为你问左连接我会做左连接。从帖子开始......

from posts p

然后让我们加入社区

left join community c on c.id_followed = p.id_user and ID_follower = 1

现在，这将为您提供所有帖子的列表以及社区id_followed记录...如果来自社区的记录为空，则来自用户joe的记录不会跟随。因为我们只想要记录乔不遵循

where c.id_followed is null

我们将为用户添加联接以获取海报名称并将其全部放在一起

from posts p
left join community c on c.id_followed = p.id_user and ID_follower = 1
left join users u on u.id_user = p.id_user
where c.id_followed is null

最后将选择行放在一起以获取所需的字段，使用concat获取您想要的名称格式

select p.id_post, p.id_user,  concat(p.post  , ' by ', u.name)

把它放在一起运行！

Answer 3

您可以在不在

中使用内部联接

select p.*
from  Posts p 
inner join (
  select id_user form community 
  where id_user  not in (
   select id_followed
   from community 
   where id_follower =1
  )
) t  on p.id_user = t.id_user and p.id_user <>1

检索用户的帖子我不遵循LeftJoin Sql

3 个答案: