Group result by matching two columns in SQL Server

问题

I am using SQL Server 2008 R2. I have a table called Messages where I store user messages each user send to other users. The table structure is like below.

+--------+----------+-----------------+------------+
| Sender | Receiver |     Message     |    Date    |
+--------+----------+-----------------+------------+
| John   | Dennis   | How are you     | 2015-06-06 |
| John   | Dennis   | Hi              | 2015-06-05 |
| Tom    | John     | How much is it? | 2015-06-04 |
| Tom    | John     | Did you buy it? | 2015-06-03 |
| Robin  | Tom      | Hey man         | 2015-06-03 |
| Dennis | John     | What up         | 2015-06-02 |
| John   | Tom      | Call me         | 2015-06-01 |
+--------+----------+-----------------+------------+

I want to get the newest message and other participants' name for a selected user for each conversation. For a example there are three conversations. One is between "john-Dennis" and 2nd one is "John-Tom"and 3rd one between "Robin-Tom".

If I want to get conversations for user john, I want to get the latest conversation message with the name of other user who is in the conversation.

The expected result for the above scenario should be like this.

+-------------+-----------------+------------+
| Participant |     Message     |    Date    |
+-------------+-----------------+------------+
| Dennis      | How are you     | 2015-06-06 |
| Tom         | How much is it? | 2015-06-04 |
+-------------+-----------------+------------+

How to achieve this using a SQL query in SQL Server. I am struggling with part for days. Please help. Thanks in advance.

回答1:

It's possible to compress this a bit, but I've split it into simple steps to hopefully make it a little easier to follow.

-- Sample data from the question.
declare @msg table (Sender varchar(32), Receiver varchar(32), [Message] varchar(max), [Date] date);
insert @msg
    (Sender, Receiver, [Message], [Date])
values
    ('John','Dennis', 'How are you', '2015-06-06'),
    ('Dennis', 'John', 'Hi', '2015-06-05'),
    ('Tom', 'John', 'How much is it?', '2015-06-04'),
    ('Tom', 'John', 'Did you buy it?', '2015-06-03'),
    ('Robin', 'Tom', 'Hey man', '2015-06-03'),
    ('Dennis', 'John', 'What up', '2015-06-02'),
    ('John', 'Tom', 'Call me', '2015-06-01');

-- The name of the user whose conversations you want to find.
declare @UserName varchar(32) = 'John';

-- Step 1: Create columns [Participant1] and [Participant2] that will be the same for
--         each pair of users regardless of who's the sender and who the receiver.
with NameOrderCTE as
(
    select 
        Participant1 = case when Sender < Receiver then Sender else Receiver end,
        Participant2 = case when Sender < Receiver then Receiver else Sender end,
        *
    from
        @msg
),

-- Step 2: For each distinct pair of participants, create a [Sequence] number that 
--         puts the messages in reverse chronological order.
MessageSequenceCTE as
(
    select
        *,
        [Sequence] = row_number() over (partition by Participant1, Participant2 order by [Date] desc)
    from
        NameOrderCTE
)

-- Step 3: Get the most recent ([Sequence] = 1) messages for each conversation
--         involving the target user.
select
    Participant = case @UserName when Sender then Receiver else Sender end,
    [Message],
    [Date]
from
    MessageSequenceCTE
where
    @UserName in (Sender, Receiver) and
    [Sequence] = 1;

回答2:

I think it should give you what you need :

SELECT a.Sender, a.Receiver, a.Message, a.[Date]
FROM 
(
   SELECT m.* , ROW_NUMBER() OVER (PARTITION BY 
   CASE 
     WHEN m.Sender = 'John' THEN 1 
     ELSE 2 
   END 
   ORDER BY [Date] DESC) AS rn
   FROM message m
   WHERE m.Sender = 'John' or m.Receiver ='John'
)a WHERE rn= 1

回答3:

Try this query

select a.Reciver as Participant,a.message,a.Date from message a
join(select sender,Receiver,max(date) from message where Sender = 'John'  group by sender,Receiver) b
on a.sender=b.sender and a.Receiver=b.Receiver and a.Date=b.date

回答4:

Base on my understanding, this do the job. Although it isn't the most performing solution because of tempDb access.

DECLARE @SelectedUser   NVARCHAR(100) = 'John'

SELECT TOP 1    Participant = Receiver,
                [Message],
                [Date]
INTO #tmp
FROM [Messages]
WHERE Sender = @SelectedUser
ORDER BY [Date]


SELECT *
FROM #tmp
UNION ALL
SELECT TOP 1 Participant = Sender,
       [Message],
       [Date]
FROM [Messages]
WHERE Receiver = @SelectedUser
ORDER BY [Date]

来源：https://stackoverflow.com/questions/31146053/group-result-by-matching-two-columns-in-sql-server