CODEWITHSUNDEEP
sqlsql-serverjoingroup-bydistinct
user1107078
user1107078

Reputation: 455

T-SQL Group By With Distinct Value

I have these tables.

User:

|ID|Field1....|

UserMessages:

|ID|UserSenderId|UserReceiverId|Message|CreatedOn|

Where UserSenderId and UserReceiverId are FK of PK User.Id

I would like to select the first message of each conversation between users (like Facebook where on the left I see a list of users and the first chars of the last message)

I tried with this query but I have multiple results for the same couple of users

SELECT DISTINCT 
    UserMessages.* 
FROM 
    UserMessages 
JOIN 
    (SELECT 
         MAX(UserMessages.Id) AS IDMAX,
         UserMessages.IdReceiver,
         UserMessages.IdSender 
     FROM 
         UserMessages 
     WHERE 
         UserMessages.IdReceiver = @UserId OR UserMessages.IdSender = @UserId 
     GROUP BY 
         UserMessages.IdReceiver, UserMessages.IdSender) IDMAX ON IDMax.IDMAX = UserMessages.Id

Example

UserMessages:

|ID|UserSenderId|UserReceiverId|Message|
  1|46|47|Hello
  2|47|46|Hello!
  3|46|48|Hey!!
  4|50|46|How are you?
  5|51|49| Hello

Considering logged userid = 46, I would like to select only the messages with id 2 3 4, where id = 2 is the last one between user 46 And 47 (logged user could be sender or receiver)

Thanks!

Upvotes: 0

Views: 797

Answers (1)

Gordon Linoff
Gordon Linoff

Reputation: 1271241

You can do this with row_number(). The trick is to pair messages regardless of who the sender and receiver are. You can do this by putting the smaller id first, then the larger id:

select um.*
from (SELECT um.*,
             row_number() over (partition by (case when idReceiver < idSender then idReceiver else idSender end),
                                             (case when idReceiver < idSender then idSender else idReceiver end)
                                order by id desc
                               ) as seqnum
     from UserMessages um
     where um.IdReceiver = @UserId OR um.IdSender = @UserId
    ) um
where seqnum = 1;

Upvotes: 2

Related Questions