Get the newest record from each group

Question

I want to sort the last messages with every user that a specific user has chated from ejabberd archive table.

The fields that I'm using are these

• id (message id)
• username (username copy)
• bare_peer (user that is chatting with)
• txt (text chat)
• created_at (time created)

What I'm trying to achieve is something like that, but I need to group message by bare_peer with username as 1_usernode, but only the last messages.

I already tested a lot of queries, but none of them worked. This is the first query I tried.

SELECT id, username, bare_peer, txt FROM archive where
username = '1_usernode' GROUP BY bare_peer ORDER BY created_at DESC;

And this is the output.

+------+------------+-------------------------------------------------------+---------------------+
| id   | username   | bare_peer              | txt                          | created_at          |
+------+------------+------------------------+------------------------------+---------------------+
| 1095 | 1_usernode | 10_usernode@localhost  | Hello !!!                    | 2016-07-17 21:15:17 |
| 1034 | 1_usernode | 15_usernode@localhost  | hey sup ?                    | 2016-07-13 22:40:29 |
| 1107 | 1_usernode | 13_usernode@localhost  | oi                           | 2016-07-18 00:09:28 |
| 1078 | 1_usernode | 2_usernode@localhost   | Hello this is just a Test!!! | 2016-07-15 16:30:50 |
| 1101 | 1_usernode | 7_usernode@localhost   | hey                          | 2016-07-18 00:05:55 |
| 1084 | 1_usernode | 3_usernode@localhost   | Hey how are you?             | 2016-07-15 19:36:44 |
| 1085 | 1_usernode | 4_usernode@localhost   | Hey how are you doing ?      | 2016-07-17 19:20:00 |

Answer 1

Try this query :-

SELECT archive.id, archive.max_id, archive.username, archive.bare_peer, archive.txt 
FROM archive join 
(SELECT MAX(id) max_id, username, bare_peer, txt 
FROM archivewhere username = '1_usernode' GROUP BY bare_peer)
tab on archive.id=tab.max_id

Answer 2

Use This Query It's Helful.

SELECT MAX(id), username, bare_peer, txt FROM archive where
username = '1_usernode' ORDER BY created_at DESC

Answer 3

declare created_at as datetime

try this

DROP TABLE IF EXISTS `archive`;

CREATE TABLE `archive` (
  `id` INT(11) NOT NULL AUTO_INCREMENT,
  `username` VARCHAR(50) DEFAULT NULL,
  `bare_peer` VARCHAR(50) DEFAULT NULL,
  `txt` TEXT,
  `created_at` DATETIME DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=INNODB  DEFAULT CHARSET=latin1;

/*Data for the table archive */

INSERT  INTO `archive`(`id`,`username`,`bare_peer`,`txt`,`created_at`) 
VALUES (1034,'1_usernode','15_usernode@localhost','hey sup ?','2016-07-13 22:40:29'),
(1078,'1_usernode','2_usernode@localhost','Hello this IS just a Test!!!','2016-07-15 16:30:50'),
(1084,'1_usernode','3_usernode@localhost','Hey how are you?','2016-07-15 19:36:44'),
(1085,'1_usernode','4_usernode@localhost','Hey how are you doing ?','2016-07-17 19:20:00'),
(1095,'1_usernode','10_usernode@localhost','Hello !!!','2016-07-17 21:15:17'),
(1101,'1_usernode','7_usernode@localhost','hey','2016-07-18 00:05:55'),
(1107,'1_usernode','13_usernode@localhost','oi','2016-07-18 00:09:28');

Then run your query

SELECT id, username, bare_peer, txt FROM archive where
username = '1_usernode' GROUP BY bare_peer ORDER BY created_at DESC;

Answer 4

Try this following code:-

select m.*
from
messages m
inner join (
select max(id) as maxid
from messages
group By (if(username > bare_peer,  username, bare_peer)), 
(if(username > bare_peer,  bare_peer, username))
) t1 on m.id=t1.maxid ;

m is alias of message table

Answer 5

You want the entry with max(created_at) for every username and bare_peer. One way to do that in MySQL is with 'having' but I don't like that. I would first get the max(created_at) for every entry:

select username, bare_peer, max(created_at) as m_
from archive
group by username, bare_peer;

Then join the table on that result:

select b.*
from (
    select username, bare_peer, max(created_at) as m_
    from archive
    group by username, bare_peer
) a
inner join archive as b on (
    a.username = b.username
    and a.bare_peer = b.bare_peer
    and a.m_ = b.created_at
)

Answer 6

I want to know why it shows the created_at column when you don't select created_at? And I don't know why you use group by? There is nothing need to be divided into groups.

My statement looks like this.select id, username, bare_peer, txt, created_at from archive where username = '1_usercode' order by created_at desc

Answer 7

I created a temporary solution with the Rahauto's answer. I put his query that returns me the correct id from latest message inside a subquery so that i can extract message content from it's id.

SELECT username, bare_peer, txt, created_at FROM archive WHERE id IN (
  SELECT tab.max_id  FROM
    archive JOIN (SELECT MAX(id) max_id, username, bare_peer, txt FROM
    archive WHERE username = '1_usernode' GROUP BY bare_peer)
  tab ON archive.id=tab.max_id
);