Select 来自按另一个 ID 分组的表的最新（前 1）

Question

I have a table that contains messages to and from users. How can I use SQL to select all of the messages that the user sent. I want to group the results by MessageSenderUserID. I also want to return only the most recent per recipient.

I tried to use MAX in a having clause, but it seems that's not the correct solution.

I may best describe it as a combination of the following queries:

SELECT TOP 1 MessageID, MessageSent, MessageSenderUserID, MessageRecipientUserID
FROM [Messaging_Message]
WHERE MessageSenderUserID = 799
ORDER BY MessageSent DESC

SELECT MessageSenderUserID
FROM [Messaging_Message]
GROUP BY MessageSenderUserID

Thanks!

Answer 1

I've only recently learned this (most surprising & fun) way of getting top 1 item in a group:

select top 1 with ties
    MessageID, 
    MessageSent, 
    MessageSenderUserID, 
    MessageRecipientUserID
from [Messaging_Message]
order by row_number() over (partition by MessageSenderUserID 
                            order by MessageSent desc)

Trick is in order by - results are grouped in partition by part and ordered by sort key, resulting in each group receiving number 1 for first row. With ties returns all ones. Oh joy of applied mechanic!

Answer 2

WITH TestTableCTE AS
(
    SELECT RN = ROW_NUMBER() OVER(PARTITION BY MessageSenderUserID ORDER BY MessageSent DESC),
           MessageID, MessageSent, MessageSenderUserID, MessageRecipientUserID
    FROM   [Messaging_Message]
)

SELECT MessageID, MessageSent, MessageSenderUserID, MessageRecipientUserID FROM TestTableCTE WHERE RN=1

Answer 3

This assumes that a user can't send more than one message at the exact same time (it would show more than one message for a user in that case):

SELECT a.MessageID, a.MessageSent, a.MessageSenderUserID, a.MessageRecipientUserID
FROM Messaging_Message AS a
    JOIN (
        SELECT MessageSenderUserID, MAX(MessageSent) AS MessageSent
        FROM Messaging_Message
        GROUP BY MessageSenderUserID
    ) AS b ON a.MessageSenderUserID = b.MessageSenderUserID
        AND a.MessageSent = b.MessageSent