獲得所有用戶的最新積分

Question

我正在獲取最新的時間戳記，但是分數比較舊。我正在尋找具有最新分數的最新時間戳記作為響應，並且每個接收者只能獲取最后一個。 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

嘗試類似：

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

這個想法是通過從tblB選擇每條記錄的最大日期來首先選擇所需的行（子查詢）。 接下來，您可以將這些記錄與原始表連接起來以獲取分數。

Answer 2

這是一般的想法。 您可以計算出細節

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須相同。