Sistema de mensajería privada. Listado del último mensaje de cada conversación

Lets say this is the database structure:

enter image description here

SELECT * FROM `pms` where id_to = 1 or id_from = 1

This would return all the messages that he has recived or sent,

So how can I retrieve the last message from cada conversation that the user 1 may have?

PD: I call it conversation when there is one or more messages between two users

operación -Editar-

So given this database content:

enter image description here

We want to get id 4 and 6

preguntado el 28 de agosto de 12 a las 14:08

3 Respuestas

Esto asume id is an auto-increment column:

SELECT MAX(id) AS id
FROM pms
WHERE id_to = 1 OR id_from = 1
GROUP BY (IF(id_to = 1, id_from, id_to))

Asumiendo que tienes id_from y id_to indexed, this variation will most likely perform better because MySQL doesn't know what to do with an OR:

SELECT MAX(id) AS id FROM
(SELECT id, id_from AS id_with
FROM pms
WHERE id_to = 1
UNION ALL
SELECT id, id_to AS id_with
FROM pms
WHERE id_from = 1) t
GROUP BY id_with

Here's how to get the messages for those ids:

SELECT * FROM pms WHERE id IN
    (SELECT MAX(id) AS id FROM
    (SELECT id, id_from AS id_with
    FROM pms
    WHERE id_to = 1
    UNION ALL
    SELECT id, id_to AS id_with
    FROM pms
    WHERE id_from = 1) t
    GROUP BY id_with)

Respondido 28 ago 12, 15:08

This seems to work! i will give it a bit of testing and let you know. thanks! - Toni Michel Caubet

@ToniMichelCaubet, another update to return the whole message. - marcus adams

I wouldn't tell.. as you can imagine I am not any master of sql. Plus, my database has very few records.. btw, I am using it ;) - Toni Michel Caubet 

select pms.* from pms 
inner join 
    (select max(fecha) as max_fecha,
           if(id_to<id_from,id_to,id_from) min_id, 
           if(id_to<id_from,id_from,id_to) max_id
      from pms where id_to = 1 or id_from = 1 
         group by if(id_to<id_from,id_to,id_from),if(id_to<id_from,id_from,id_to)) t 
     on (if(pms.id_to<pms.id_from,pms.id_to,pms.id_from)=t.min_id) 
        and (if(pms.id_to<pms.id_from,pms.id_from,pms.id_to)=t.max_id) 
        and (pms.fecha=t.max_fecha)

Also if id_to and id_from in your table are small enough to prevent overflow in statement (id_to+id_from) here is the simple query:

select pms.* from pms 
inner join 
    (select max(fecha) as max_fecha, id_to+id_from as sum_id
      from pms where id_to = 1 or id_from = 1 
         group by id_to+id_from) t 
     on ((pms.id_to+pms.id_from)=t.sum_id) 
        and (pms.fecha=t.max_fecha)
 where pms.id_to = 1 or pms.id_from = 1

Respondido 28 ago 12, 17:08

This is showing the last message of each user of each conversation. Its close, but we only want the last message of each conversation - Toni Michel Caubet

@ToniMichelCaubet, is this really the one you're going to use? - marcus adams

yours performe better? the only reason is cause i have to add all the other rows to your query and rename the max(id) as id... @MarcusAdams - Toni Michel Caubet

Works OK! But replace "GROUP BY id_to+id_from" with: "GROUP BY sum_id". And this is redundant: "((pms.id_to+pms.id_from)=t.sum_id)", also "WHERE pms.id_to = 1 or pms.id_from = 1" is a redundancy; remove that as well and the code will be much shorter / efficient. - andreszs

Esta consulta debería funcionar:

SELECT a.*
FROM pms a
     INNER JOIN (
                 SELECT id_to, id_from, MAX(fecha) AS fecha
                 FROM pms
                 WHERE (id_to = 1 OR id_from = 1)
                 GROUP BY LEAST(id_to, id_from)
                ) b
                ON a.fecha = b.fecha AND
                   (a.id_to = b.id_to OR
                   a.id_from = b.id_from);

See example @ sqlfiddle here

Si tiene id as PRIMARY KEY and you are logging messages in a chronological order, then it can be further optimized and simplified as:

SELECT a.*
FROM pms a
     INNER JOIN (
                 SELECT MAX(id) AS id
                 FROM pms
                 WHERE (id_to = 1 OR id_from = 1)
                 GROUP BY LEAST(id_to, id_from)
                ) b
                ON a.id = b.id;

Respondido 28 ago 12, 19:08

The following errors were reported:Every derived table must have its own alias - Toni Michel Caubet

ohh ok.. forgot alis for derived table b. updated query, try now. - Omesh

gosh... The following errors were reported:Column 'id_to' in where clause is ambiguous - Toni Michel Caubet

Now works but is giving the last message of each user of each conversation. Its close, but we only want the last message of each conversation - Toni Michel Caubet

The thing is that this will give 2 messages per conversation, i want only the last one.. - Toni Michel Caubet

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