私人消息傳遞系統。清單每個對話

Question

比方說，這是數據庫結構的最後一條消息：

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

這將返回所有他recived或發送的消息，

所以，我該怎麼找回來自的最後一條消息，每個用戶1可能擁有的對話？

PD：我把它叫做談話時，有兩個用戶

CNC中

因此，考慮該數據庫的內容之間的一個或多個消息：

我們希望得到ID 4和6

Answer 1

這是假設id是一個自動遞增列：

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)) 

假設你有id_from和id_to索引，這種變化很可能會表現得更好，因爲MySQL不知道做什麼用的或做的事：

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 

下面是如何獲得這些消息的這些ID：

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) 

Answer 2

此查詢應該工作：

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

如果有id如PRIMARY KEY和你正在按時間順序記錄消息，那麼它可以被進一步優化和簡化爲：

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; 

Answer 3

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) 

此外，如果id_to和您的表中的id_from足夠小，以防止語句溢出（id_to + id_from），這裏是簡單查詢：

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