Question

这似乎是一个非常简单的问题，但我无法解决它，假设我有下表：

ID  MessageId   ReceiverId
----------------------------
1   1           1
2   1           2
3   2           3
4   3           1
5   3           2

我想在MessageId列中有一个唯一值的结果，这意味着MessageId的只有一行是1，无论哪一行。像这样的东西：

ID  MessageId   ReceiverId
----------------------------
1   1           1
3   2           3
4   3           1

如果我只需要返回一列，那么我将不使用GROUP BY。

Answer 1

似乎您需要 correlated 子查询：

select m.*
from msg m
where id = (select top (1) id
            from msg m1 
            where m1.messageid = m.messageid
            order by m1.ReceiverId 
           );

Answer 2

select id, messageId, receiverId
from
(
    select *,
           row_number() over(partition by messageId order by id) rnum
    from @t
) x
where x.rnum = 1;

Answer 3

为了获得最佳性能，需要使用CTE。让我们首先使用脚本创建表：

 CREATE TABLE MyMessages(ID INT PRIMARY KEY ,  MessageId INT,   ReceiverId INT )
 GO 

INSERT INTO MyMessages values(1 ,  1 ,  1)
INSERT INTO MyMessages values(2 ,  1 ,  2)
INSERT INTO MyMessages values(3 ,  2 ,  3)
INSERT INTO MyMessages values(4 ,  3 ,  1)
INSERT INTO MyMessages values(5 ,  3 ,  2)

我认为该查询对您来说很好。

;WITH UniqueMessages AS (SELECT MIN(ID) AS FirstID FROM MyMessages AS MM GROUP BY MM.MessageId)

SELECT  MM.ID, MM.MessageId, MM.ReceiverId 
FROM UniqueMessages AS UM 
INNER JOIN MyMessages AS MM ON MM.ID = UM.FirstID

Answer 4

您可以将CTE与RANK一起使用

CREATE TABLE #test(ID INT, MessageId INT,  ReceiverId INT)
INSERT INTO #test VALUES
(1,1,1),
(2,1,2),
(3,2,3),
(4,3,1),
(5,3,2)

WITH val AS (
    SELECT *, RANK() OVER(PARTITION BY MessageId ORDER BY ID) rnk
    FROM #test
    )
SELECT ID, MessageId, ReceiverId 
FROM val WHERE rnk = 1

输出：

ID  MessageId   ReceiverId
1   1           1
3   2           3
4   3           1

Answer 5

SELECT *
FROM #myTable
WHERE ID IN
      (
          SELECT MIN(ID) FROM #myTable GROUP BY MessageId
      );

Answer 6

假设一行只能出现两次，我建议：

select m.*
from msgs m
where MessageId < ReceiverId
union all
select m.*
from msgs m
where MessageId > ReceiverId and
      not exists (select 1
                  from msgs m2
                  where m2.MessageId = m.ReceiverId and
                        m2.ReceiverId = m.MessageId
                 );

如果确实有两行以上，则可以使用不需要子查询的此解决方案：

select top (1) with ties m.*
from msgs m
order by row_number() over (order by case when MessageId < ReceiverId then MessageId 
                                           else ReceiverId
                                      end
                           );

从一列中只为另一个列中的每个值选择一个值，以及对应的其他列值

6 个答案: