Question

这是我的对话表：

conversationID  ||  userID
      1         ||    1
      1         ||    2
      2         ||    1
      2         ||    2
      2         ||    3

您可以看到每个会话可以包含2个或更多用户。

我试图获得只有2个用户在那里的对话的ID。即仅包含用户1和1的会话。 2，答案是对话1。

但我怎么得到它？

Answer 1

你应该使用having子句。假设（conversationID，userID）是PK或AK，您的查询是：

  select   conversationID
    from   your_Table
group by   conversationID
  having   count( * ) = 2

已编辑加入了1,2个用户对话，这是一种索引友好的方法，没有相关的子查询，也没有逐行的功能。

      select   t1 conversationID
        from   your_Table t1
  inner join 
               ( select distinct conversationID
                  from your_Table
                 where userId in (1, 2)
               ) t2
          on   t1.conversationID = t2.conversationID
    group by   t1.conversationID
      having   count( distinct t1.userId ) = 2

Answer 2

这将选择所有具有用户1或用户2或两者的会话，但不会选择其他人：

select conversationID
from conversations
group by conversationID
having count(*) = count(case when userID in (1,2) then 1 end)

如果您还希望所有具有完全用户1和2的会话，而不是其他人，则还必须添加和条件：

select conversationID
from conversations
group by conversationID
having count(*) = count(case when userID in (1,2) then 1 end)
       and count(*) = 2 -- number of elements in set

如果可以复制userID，最好使用distinct：

select conversationID
from conversations
group by conversationID
having
  count(distinct userID) = count(distinct case when userID in (1,2) then userID end)
  and count(distinct userID) = 2 -- number of elements in set

Answer 3

希望这会对你有所帮助，

select conversationID from conversation
group by ConversationID having count(distinct UserID)=2;

sqlfiddle demo

sql消息表查询

3 个答案: