选择SQL Server中每个唯一组合的最后一行

Question

我需要选择每个唯一组合的最后一行，而不考虑组合元素的顺序。 例如，组合(5,6)和(6,5)将被视为相同的组合。

这是表格上的数据：

我的预期输出：

到目前为止，我所做的是：

Select * From
(
Select *, ROW_NUMBER() over (Partition by SenderId, ReceiverId order by MessageId desc) as Seq From Messages
) as TempTalbe Where Seq = 1

我的查询的输出是：

SQL Server专家有什么帮助吗？

Answer 1

您可以为此使用row_number() 。 传统方法是：

select m.*
from (select m.*,
             row_number() over (partition by (case when senderId < receiverId then senderId else receiverId end),
                                             (case when senderId < receiverId then receiverId else senderId end)
                                order by messageSentAt desc
                               ) as seqnum
      from messages m
     ) m
where seqnum = 1;

您也可以在不使用子查询的情况下对此进行措辞：

select top (1) with ties m.*
from messages m
order by row_number() over (partition by (case when senderId < receiverId then senderId else receiverId end),
                                         (case when senderId < receiverId then receiverId else senderId end)
                            order by messageSentAt desc
                           );

Answer 2

尝试这个：

select messageId,
       messageBody,
       messageSentAt,
       isRead
from (
    select messageId,
           messageBody,
           messageSentAt,
           isRead,
           row_number() over (partition by senderId, receiverId order by messageSentAt desc) rn
    from (
        select messageId,
               case when senderId > receiverId then senderId else receiverId end senderId,
               case when senderId < receiverId then senderId else receiverId end receiverId,
               messageBody,
               messageSentAt,
               isRead
        from Messages
    ) a
) a where rn = 1

最内部的查询统一了发送者和接收者，因此它们的顺序无关紧要：

case when senderId > receiverId then senderId else receiverId end senderId,
case when senderId < receiverId then senderId else receiverId end receiverId,

中间查询为每行分配行号，因此第一个是最新日期。 大多数外部查询只是根据该行号进行过滤。