获得所有用户的最新积分

Question

我正在获取最新的时间戳记，但是分数比较旧。我正在寻找具有最新分数的最新时间戳记作为响应，并且每个接收者只能获取最后一个。 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;

Answer 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

这个想法是通过从tblB选择每条记录的最大日期来首先选择所需的行（子查询）。 接下来，您可以将这些记录与原始表连接起来以获取分数。

Answer 2

这是一般的想法。 您可以计算出细节

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

您看到的地方在任何地方都必须相同。