查找在一列中具有相同值而在另一列中具有其他值的行？

Question

我有一个 PostgreSQL 数据库，它将用户存储在 users table 中，他们参与的对话存储在 conversation table 中。由于每个用户都可以参与多个对话，并且每个对话都可以涉及多个用户，因此我有一个 conversation_user 链接 table 来跟踪哪些用户参与了每个对话：

# conversation_user
id  |  conversation_id | user_id
----+------------------+--------
1   |                1 |      32
2   |                1 |       3
3   |                2 |      32
4   |                2 |       3
5   |                2 |       4

在上面的 table 中，用户 32 仅与用户 3 进行了一次对话，而另一次与用户 3 和用户 4 进行了一次对话。我将如何编写一个查询来显示只有用户之间存在对话32 和用户 3？

我试过以下方法：

SELECT conversation_id AS cid,
       user_id
FROM conversation_user
GROUP BY cid HAVING count(*) = 2
AND (user_id = 32
     OR user_id = 3);

SELECT conversation_id AS cid,
   user_id
FROM conversation_user
GROUP BY (cid HAVING count(*) = 2
AND (user_id = 32
     OR user_id = 3));

SELECT conversation_id AS cid,
       user_id
FROM conversation_user
WHERE (user_id = 32)
  OR (user_id = 3)
GROUP BY cid HAVING count(*) = 2;

这些查询抛出一个错误，表明 user_id 必须出现在 GROUP BY 子句中或用于聚合函数。将它们放在聚合函数中（例如 MIN 或 MAX）听起来不合适。我认为我的前两次尝试是将它们放在 GROUP BY 子句中。

我做错了什么？

Answer 1

这是一个关系划分的案例。我们在这个相关问题下汇集了一系列技术：

How to filter SQL results in a has-many-through relation

特别困难的是排除额外的用户。基本上有4个技巧。

Select rows which are not present in other table

我建议 LEFT JOIN / IS NULL:

SELECT cu1.conversation_id
FROM        conversation_user cu1
JOIN        conversation_user cu2 USING (conversation_id)
LEFT   JOIN conversation_user cu3 ON cu3.conversation_id = cu1.conversation_id
                                 AND cu3.user_id NOT IN (3,32)
WHERE  cu1.user_id = 32
AND    cu2.user_id = 3
AND    cu3.conversation_id IS NULL;

或NOT EXISTS:

SELECT cu1.conversation_id
FROM   conversation_user cu1
JOIN   conversation_user cu2 USING (conversation_id)
WHERE  cu1.user_id = 32
AND    cu2.user_id = 3
AND NOT EXISTS (
   SELECT 1
   FROM   conversation_user cu3
   WHERE  cu3.conversation_id = cu1.conversation_id
   AND    cu3.user_id NOT IN (3,32)
   );

这两个查询不取决于 (conversation_id, user_id) 的 UNIQUE 约束，该约束可能存在也可能不存在。这意味着，如果 user_id 32（或 3）在同一个对话中被多次列出，查询甚至会起作用。但是，您会在结果中得到重复的行，并且需要应用 DISTINCT 或 GROUP BY.
唯一的条件是你制定的条件：

... a query that would show that there is a conversation between just user 32 and user 3?

审核查询

query you linked in the comment 行不通。您忘记排除其他参与者。应该是这样的：

SELECT *  -- or whatever you want to return
FROM   conversation_user cu1
WHERE  cu1.user_id = 32
AND    EXISTS (
   SELECT 1
   FROM   conversation_user cu2
   WHERE  cu2.conversation_id = cu1.conversation_id 
   AND    cu2.user_id = 3
   )
AND NOT EXISTS (
   SELECT 1
   FROM   conversation_user cu3
   WHERE  cu3.conversation_id = cu1.conversation_id
   AND    cu3.user_id NOT IN (3,32)
   );

这与其他两个查询类似，只是如果 user_id = 3 被多次链接，它不会 return 多行。

Answer 2

您可以使用条件聚合来 select 所有只有 2 个特定参与者的 cid

select cid from conversation_user
group by cid
having count(*) = 2
and count(case when user_id not in (32,3) then 1 end) = 0

如果 (cid,user_id) 不唯一，则将 having count(*) = 2 替换为 having count(distinct user_id) = 2

Answer 3

因为您只想与 2 个用户进行对话，您可以对其他用户使用自外连接并过滤掉命中：

要查找所有 2 用户对话，并且它们介于：

SELECT
    a.conversation_id cid,
    a.user_id user_id_1,
    b.user_id user_id_2
FROM conversation_user a
JOIN conversation_user b ON b.cid = a.cid
  AND b.user_id > a.user_id
LEFT JOIN conversation_user c ON c.cid = a.cid
  AND c.user_id NOT IN (a.user_id, b.user_id)
WHERE c.cid IS NULL -- only return misses on join to others

要查找特定用户的所有 2 用户对话，只需添加：

AND a.user_id = 32

Answer 4

如果你只是想确认。

 select conversation_id 
   from  conversation_users 
   group by conversation_id
   having bool_and ( user_id in (3,32))
      and count(*) = 2;

如果你想要完整的细节，您可以使用 window 函数和 CTE，如下所示：

 with a as (
   select *
      ,not bool_and( user_id in (3,32) )
         over  ( partition by conversation_id) 
       and 2 = count(user_id)
         over  ( partition by conversation_id)
           as conv_candidates 
   from conversation_users 
   ) 
 select * from a where conv_candidates;

查找在一列中具有相同值而在另一列中具有其他值的行？

Find rows that have same value in one column and other values in another column?

sql

postgresql

relational-division

审核查询