SQL查询最新消息 [英] SQL query for most recent messages

问题描述

我正在尝试实现一个非常类似于facebook的消息系统.消息表是:

I am trying to implement a message system quite similar to facebook . The message table is :

+--------+----------+--------+-----+----------+
| msg_id | msg_from | msg_to | msg | msg_time |
+--------+----------+--------+-----+----------+

此处msg_from和msg_to包含用户ID，而msg_time包含消息的时间戳.一个用户的用户ID可以同时出现在至"和来自"列中，而另一个用户可以多次出现.我应该如何编写一个SQL查询来选择两个用户之间最近发送的消息? (消息可以来自任何一个)1到2或2到1.

Here msg_from and msg_to contain user ids and the msg_time contains the timestamp of the message . A user's user id can appear in both the to and from column and multiple times for another user . How should I write a SQL query which selects the most recent sent message between two users ? (The message can come from either one) 1 to 2 or 2 to 1 .

推荐答案

由于John Woo明确指出它不是定向的，所以这是我的新答案:

Since John Woo clarified that it is not directional, here's my new answer:

select *
from msgsList
where (least(msg_from, msg_to), greatest(msg_from, msg_to), msg_time)       
in 
(
    select 
       least(msg_from, msg_to) as x, greatest(msg_from, msg_to) as y, 
       max(msg_time) as msg_time
    from msgsList 
    group by x, y
);

输出:

| MSG_ID | MSG_FROM | MSG_TO |    MSG |                       MSG_TIME |
------------------------------------------------------------------------
|      1 |        1 |      2 |  hello | January, 23 2010 17:00:00-0800 |
|      5 |        1 |      3 | me too | January, 23 2012 00:15:00-0800 |
|      6 |        3 |      2 |  hello | January, 23 2012 01:12:12-0800 |

对于此输入:

create table msgsList
(
  msg_id int,
  msg_from int, 
  msg_to int,
  msg varchar(10),
  msg_time datetime
);

insert into msgslist VALUES

(1, 1, 2, 'hello', '2010-01-23 17:00:00'),      -- shown
(2, 2, 1, 'world', '2010-01-23 16:00:00'),

(3, 3, 1, 'i am alive', '2011-01-23 00:00:00'),
(4, 3, 1, 'really', '2011-01-22 23:15:00'),
(5, 1, 3, 'me too', '2012-01-23 00:15:00'),     -- shown

(6, 3, 2, 'hello', '2012-01-23 01:12:12');      -- shown

SQLFiddle演示

---

如果您最喜欢使用ANSI SQL，请按以下方法操作: http://sqlfiddle.com/#!2/0a575/19

SQLFiddle Demo

---

If ANSI SQL is your cup of tea, here's the way to do it: http://sqlfiddle.com/#!2/0a575/19

select *
from msgsList z
where exists
(
    select null
    from msgsList
    where 
      least(z.msg_from, z.msg_to) = least(msg_from, msg_to)
      and greatest(z.msg_from, z.msg_to) = greatest(msg_from, msg_to)
    group by least(msg_from, msg_to), greatest(msg_from, msg_to)
    having max(msg_time) = z.msg_time  
) ;

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