用户号码= 1,即Aa
:
CREATE TABLE if not exists tblA
(
id int(11) NOT NULL auto_increment ,
sender varchar(255),
receiver varchar(255),
msg varchar(255),
date timestamp,
PRIMARY KEY (id)
);
CREATE TABLE if not exists tblB
(
id int(11) NOT NULL auto_increment ,
sno varchar(255),
name varchar(255),
PRIMARY KEY (id)
);
INSERT INTO tblA (sender, receiver,msg,date ) VALUES
('1', '2', 'buzz ...','2011-08-21 14:11:09'),
('1', '2', 'test ...','2011-08-21 14:12:19'),
('1', '2', 'check ...','2011-08-21 14:13:29'),
('1', '1', 'test2 ...','2011-08-21 14:14:09'),
('2', '1', 'check2 ...','2011-08-21 14:15:09'),
('2', '1', 'test3 ...','2011-08-21 14:16:09'),
('1', '2', 'buzz ...','2011-08-21 14:17:09'),
('1', '2', 'test ...','2011-08-21 14:18:19'),
('1', '2', 'check ...','2011-08-21 15:19:29'),
('1', '1', 'test2 ...','2011-08-21 14:10:09'),
('3', '1', 'check2 ...','2011-08-21 14:21:09'),
('3', '1', 'test3 ...','2011-08-21 14:22:09'),
('3', '2', 'buzz ...','2011-08-21 14:24:09'),
('3', '2', 'test ...','2011-08-21 14:25:19'),
('1', '3', 'check ...','2011-08-21 14:26:29'),
('1', '3', 'test2 ...','2011-08-21 14:27:09'),
('2', '3', 'check2 ...','2011-08-21 14:28:09'),
('2', '3', 'test3 ...','2011-08-21 14:29:09'),
('1', '2', 'check3 ...','2011-08-21 14:23:09'),
('1', '4', 'test2 ...','2011-08-21 14:27:09'),
('1', '5', 'test2 ...','2011-08-21 14:27:09'),
('2', '6', 'check2 ...','2011-08-21 14:28:09'),
('1', '7', 'test3 ...','2011-08-21 14:29:09'),
('8', '2', 'check3 ...','2011-08-21 14:23:09');
INSERT INTO tblB (sno, name ) VALUES
('1', 'Aa'),
('2', 'Bb'),
('3', 'Cc'),
('4', 'Dd'),
('5', 'Ee'),
('6', 'Ff'),
('7', 'Gg'),
('8', 'Hh');
如何获得最新通信时间b / n 2个用户。 http://www.sqlfiddle.com/#!2/ed676/1查询未提供最新的通信时间。
我非常感谢任何帮助。谢谢你。
答案 0 :(得分:1)
这是你想要的吗?
select b.name, max(date)
from tblA a join
tblB b
on b.sno in (a.receiver, a.sender)
where b.sno <> '1' and
exists (select 1
from tblB b2
where b2.sno = '1' and
b2.sno in (a.receiver, a.sender)
) or
(a.sender = 1 and a.receiver = 1)
group by b.name
order by max(date) desc;
这将返回与1
交谈的每个“其他人”的最新通信时间。您的原始查询未将max(date)
放入选择列表中。
答案 1 :(得分:0)
您最好的选择可能是使用UNION
,例如:
select * from (
select * from tblA
where sender = '1' and receiver = '2'
union
select * from tblA
where sender = '2' and receiver = '1'
) as t
order by date desc
limit 5
你可以做一些事情来提高效率,比如降低一些限制并排序到UNION
的每个部分,但这是基本的想法。
答案 2 :(得分:0)
根据我之前回答的讨论,我认为你想要的是这样的:
SELECT
tblA.id,
sender.name AS sender_name,
receiver.name AS receiver_name,
tblA.msg,
tblA.date
FROM (
SELECT
LEAST(sender, receiver) AS a,
GREATEST(sender, receiver) AS b,
MAX(id) AS id
FROM tblA
GROUP BY a, b
) AS t
JOIN tblA ON t.id = tblA.id
JOIN tblB AS sender ON tblA.sender = sender.sno
JOIN tblB AS receiver ON tblA.receiver = receiver.sno
所以基本的想法是:
id
)表选择最新消息auto_increment
,PRIMARY KEY
或其他形式的连续递增数字tblA
。这会查找这些消息的ID,而不考虑sender
或receiver
排序。id
,选择其详细信息。tblB
)记录加入消息行两次,一次获取sender
详细信息,一次获取receiver
详细信息。答案 3 :(得分:0)
试试这个:
select p1.name,p2.name,max(date) from
(select least(sender,receiver) min_val,greatest(sender,receiver) max_val,msg,date from tblA
order by min_val,max_val,date) t inner join tblB p1 on t.min_val=p1.sno
inner join tblB p2 on t.max_val=p2.sno
group by min_val,max_val;