我正在获取最新的时间戳,但是得分很高。我正在寻找最新的时间戳,最新的分数作为响应,每个接收器只有最后一个。 http://www.sqlfiddle.com/#!2/07d11/1
CREATE TABLE if not exists tblA
(
id int(11) NOT NULL auto_increment ,
sender varchar(255),
receiver varchar(255),
msg varchar(255),
date timestamp,
points varchar(255),
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,points ) VALUES
('1', '2', 'buzz ...','2011-08-21 14:11:09','10'),
('1', '2', 'test ...','2011-08-21 14:12:19','20'),
('1', '3', 'buzz ...','2011-08-21 14:11:09','10'),
('1', '3', 'test ...','2011-08-21 14:12:19','20'),
('1', '4', 'buzz ...','2011-08-21 14:11:09','10'),
('1', '4', 'test ...','2011-08-21 14:12:19','20');
INSERT INTO tblB (sno, name ) VALUES
('1', 'Aa'),
('2', 'Bb'),
('3', 'Cc'),
('4', 'Dd'),
('5', 'Ee'),
('6', 'Ff'),
('7', 'Gg'),
('8', 'Hh');
sql:
select *, max(date)
from tblA a join
tblB b
on b.sno in (a.receiver)
group by b.name
order by max(date) desc;
答案 0 :(得分:1)
尝试类似:
SELECT *
FROM (
SELECT tblB.*, MAX(tblA.date) AS date
FROM tblB
JOIN tblA ON tblB.sno = tblA.receiver
GROUP BY tblB.sno
) AS subset
JOIN tblA ON subset.sno = tblA.receiver
AND subset.date = tblA.date
我们的想法是首先选择所需的行(子查询),方法是从tblB
中选择每条记录的最大日期。接下来,您可以将这些记录与原始表联系起来以获取分数。
答案 1 :(得分:0)
这是一般性的想法。你可以弄清楚细节
select field1, field2, etc
from table
join (
select id, max(timestamp) maxts
from table
where whatever
group by id ) temp on table.id = temp.id
where whatever
and timestamp = ts
无论哪个部分都必须与你看到的地方相同。