Question

我有一个非常简单的表，其中包含几列：

ID           (int)
Sender ID    (int)
Recipient ID (int)
Date Time    (DateTime)
MessageText  (varchar(255))

问题是，我想从此表中检索发件人ID或收件人ID等于我的用户ID的最新记录，但我还希望其他列是唯一的。

让我举个例子，我有一些行包含以下数据：

    ID    Sender   Recipient          Date              Message
    1       1         2       2011-01-01 01:55:00    Test Data
    2       1         2       2011-01-01 01:56:00    Test Data 2
    3       2         1       2011-01-01 01:59:00    Some more test data
    4       3         1       2011-01-02 11:50:00    Test Data 3

我希望我的查询/存储过程在此数据上返回2行。 ID为3的行和ID为4的行，但我似乎无法在单个存储过程中解决这个问题。

编辑：发现它，如果有人有任何优化，请告诉我

ALTER PROCEDURE GetRootMessages
    @UserID         int
AS
BEGIN
    DECLARE @OtherUserID int

    CREATE TABLE #TempMessages 
    (
        ID int, 
        SenderID int, 
        RecipientID int, 
        MessageText text, 
        SendDate datetime not null, 
        ReadDate datetime null, 
        ReplyTo int null);

    DECLARE idCursor    Cursor FOR
    (
        SELECT DISTINCT SenderID as OtherUserID FROM Messages WHERE RecipientID = @UserID
        UNION
        SELECT DISTINCT RecipientID as OtherUserID FROM Messages WHERE SenderID = @UserID
    );

    OPEN idCursor
    FETCH NEXT FROM idCursor INTO @OtherUserID
    WHILE @@FETCH_STATUS = 0
    BEGIN
        INSERT INTO #TempMessages (ID, SenderID, RecipientID, MessageText, SendDate, ReadDate, ReplyTo)
        SELECT TOP 1 * 
        FROM Messages 
        WHERE (SenderID = @UserID AND RecipientID = @OtherUserID) OR (SenderID = @OtherUserID AND RecipientID = @UserID)
        ORDER BY SendDate DESC

        FETCH NEXT FROM idCursor INTO @OtherUserID
    END
    CLOSE idCursor
    DEALLOCATE idCursor

    SELECT * FROM #TempMessages ORDER BY SendDate DESC
    DROP TABLE #TempMessages
END

Answer 1

在SQL Server 2005 +中：

WITH ordered AS (
  SELECT
    *,
    Person1 = CASE WHEN Sender > Recipient THEN Recipient ELSE Sender END,
    Person2 = CASE WHEN Sender > Recipient THEN Sender ELSE Recipient END
  FROM atable
),
ranked AS (
  SELECT
    *,
    rank = ROW_NUMBER() OVER (PARTITION BY Person1, Person2 ORDER BY Date DESC)
  FROM ordered
)
SELECT
  ID,
  Sender,
  Recipient,
  Date,
  Message
FROM ranked
WHERE rank = 1

Answer 2

我认为这就是你要找的东西。

;WITH tab (row, ID, Sender, Recipient, Date, Message)
AS (select row_number() over (order by date desc), * 
  from table
 where ([Sender ID] = @id or [Recipient ID] = @id)
   and ID <> [Sender ID]
   and ID <> [Recipient ID]
   and [Sender ID] <> [Recipient ID])
SELECT ID, Sender, Recipient, Date, Message
  FROM tab 
 WHERE row = 1

基本上，您会获得用户ID的所有记录列表，并按日期排序。然后可以将第一个作为最新的一个。

在我用来重现的完整脚本下面。创建表邮件（ ID int，发件人int，收件人int， [日期]日期时间， MessageText varchar（255））

insert into mails values
    (1 ,      1  ,       2 ,      '2011-01-01 01:55:00' ,   'Test Data'),
    (2 ,      1  ,       2 ,      '2011-01-01 01:56:00' ,   'Test Data 2'),
    (3 ,      2  ,       1 ,      '2011-01-01 01:59:00' ,   'Some more test data'),
    (4 ,      3   ,      1 ,      '2011-01-02 11:50:00' ,   'Test Data 3')

    declare @id int
    select @id = 1

;WITH tab (row, ID, Sender, Recipient, [Date], Message)
AS (select row_number() over (order by [Date] desc), * 
  from mails
 where ([Sender] = @id or [Recipient] = @id)
   and ID <> [Sender]
   and ID <> [Recipient]
   and [Sender] <> [Recipient])
SELECT ID, Sender, Recipient, [Date], Message
  FROM tab 
 WHERE row = 1

要获取按最新排序的消息列表，您可以使用此查询：

select * 
  from mails
 where ([Sender] = @id or [Recipient] = @id)
   and ID <> [Sender]
   and ID <> [Recipient]
   and [Sender] <> [Recipient]
 order by [Date] desc

Answer 3

假设我没有遗漏某些东西你不能这样做吗？哪个X是你想要找的ID？

select * 
 from TABLE
where (SENDER=x and RECIPIENT<>x) 
   OR (SENDER<>x and RECIPIENT=x);

或获得最新的

SELECT * 
  FROM TABLE 
 WHERE SENDER=X 
   AND RECIPIENT <> X
   AND DATE = (SELECT MAX(DATE) 
                 FROM TABLE 
                WHERE SENDER=X)
UNION
SELECT * 
  FROM TABLE 
 WHERE SENDER <> X 
   AND RECIPIENT = X 
   AND DATE = (SELECT MAX(DATE) 
                 FROM TABLE 
                WHERE RECIPIENT=X)

Answer 4

版本1：

SELECT Sender, Recipient
FROM Messages
GROUP BY Sender, Recipient
HAVING COUNT(*) = 1

这将为您提供唯一的发件人和收件人对的列表，但您需要将其用作子查询或作为联接的一部分来获取唯一邮件列表。

第2版：

SELECT *
FROM Messages O
WHERE NOT EXIST
(SELECT *
 FROM Messages I
 WHERE I.ID <> O.ID AND I.Sender = O.Sender AND I.Recipient = O.Recipient)

也可以将其写为左外连接，但我不想尝试。

Answer 5

我认为ammianus走在正确的轨道上。

基本上你想要使用DISTINCT只获得唯一的，而UNION要结合你将得到的两个查询。

Answer 6

试试这个：

declare @UserID int
select @UserID = 1

-- Union all messages in one table and filter them on our userid
declare @Union table(ID int, UserID int, SendDate datetime, Message nvarchar(50))
insert into @Union(ID, UserID, SendDate, Message)
select ID, RecipientID, SendDate, Message
from Messages
where SenderID = @UserID
union
select ID, SenderID, SendDate, Message
from Messages
where RecipientID = @UserID

    -- Gets max dates for every correspond user
declare @MaxDates table(UserID int, SendDate datetime)
insert into @MaxDates(UserID, SendDate)
select UserID, max(SendDate)
from @Union
group by UserID

    -- Gets result
select u.*
from @Union u
    inner join @MaxDates md on (u.UserID = md.UserID) and (u.SendDate = md.SendDate)

两列中任何一列中的唯一值

6 个答案: