排名用户由最大id和点，但（排名是错误的）

Question

我想获得最高点行与行列和选择查询没有rownum = rownum + 1。我已经尝试了下面的查询，但我缺少的东西也链接http://sqlfiddle.com/#!2/fd7897/7。

我在寻找每个接收者的最后一个入口的答案，这也是最高分入口rankwise：我真的很感激任何帮助。感谢提前。

像这样：

（ '2'， '4'， '测试...'， '2011-08-21 14时13分19秒'， '40'， '009'） - --rank1

（'4'，'2'，'test ...'，'2011-08-21 14:13:19'，'30'，'003'）---- rank2

（ '1'， '3'， '测试...'， '2011-08-21 14点12分19秒'， '20'， '005'）---- rank3

查询到目前为止：

SELECT * , 
      (select count(*) 
      from tblA u2 
      where u2.points > u.points or 
        u2.points = u.points and u2.id <= u.id 
      ) as rank 
FROM (SELECT u.receiver, MAX(u.id) AS id 
     FROM tblA u 
     GROUP BY u.receiver 
    ) subset JOIN 
    tblA u 
    ON subset.receiver = u.receiver AND subset.id = u.id order by rank; 

个

表：

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 int(255), 
    refno 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) 
); 

CREATE TABLE if not exists tblC 
(
id int(11) NOT NULL auto_increment , 
data varchar(255), 
    refno varchar(255), 
    extrarefno varchar(255), 
PRIMARY KEY (id) 
); 


INSERT INTO tblA (sender, receiver,msg,date,points,refno) VALUES 
('1', '2', 'buzz ...','2011-08-21 14:11:09','10','001'), 
('1', '2', 'test ...','2011-08-21 14:12:19','20','002'), 
('4', '2', 'test ...','2011-08-21 14:13:19','30','003'), 
('1', '3', 'buzz ...','2011-08-21 14:11:09','10','004'), 
('1', '3', 'test ...','2011-08-21 14:12:19','20','005'), 
('1', '4', 'buzz ...','2011-08-21 14:11:09','10','006'), 
('1', '4', 'test ...','2011-08-21 14:12:19','20','007'), 
('3', '4', 'test ...','2011-08-21 14:13:19','30','008'), 
('2', '4', 'test ...','2011-08-21 14:13:19','40','009'); 



INSERT INTO tblB (sno, name) VALUES 
('1', 'Aa'), 
('2', 'Bb'), 
('3', 'Cc'), 
('4', 'Dd'), 
('5', 'Ee'), 
('6', 'Ff'), 
('7', 'Gg'), 
('8', 'Hh'); 


INSERT INTO tblC (data,refno,extrarefno) VALUES 
('data1', '001', '101'), 
('data2', '002', '102'), 
('data3', '003', '103'), 
('data4', '004', '101'), 
('data5', '005', '102'), 
('data6', '006', '103'), 
('data7', '007', '101'), 
('data8', '008', '101'), 
('data9', '009', '101'); 

Answer 1

问题是子查询中的count(*)。将其更改为count(distinct receiver)：

SELECT * , 
     (select count(distinct receiver) 
     from tblA u2 
     where u2.points > u.points or 
       u2.points = u.points and u2.id <= u.id 
     ) as rank 
FROM (SELECT u.receiver, MAX(u.id) AS id 
     FROM tblA u 
     GROUP BY u.receiver 
    ) subset JOIN 
    tblA u 
    ON subset.receiver = u.receiver AND subset.id = u.id 
order by rank; 

编辑：

要创建这个作为MySQL的观点，你必须得到from子句中的聚集权：

SELECT * , 
     (select count(distinct receiver) 
     from tblA u2 
     where u2.points > u.points or 
       u2.points = u.points and u2.id <= u.id 
     ) as rank 
FROM tblA u 
WHERE u.id = (select max(u2.id) from tblA u2 where u2.receiver = u.receiver) 
order by rank; 

Answer 2

怎么样？

SELECT a.* 
FROM tbla a, 
     (SELECT receiver, 
       Max(points) AS m 
     FROM tbla 
     GROUP BY receiver) AS b 
WHERE a.receiver = b.receiver 
     AND a.points = b.m 
ORDER BY m DESC  

Answer 3

请注意，相关MySQL的子查询性能较差。我认为下面的查询返回结果与您的结果相同并且速度很快。

SELECT x.*, @ord := @ord + 1 AS rank 
FROM (
    SELECT u.* 
    FROM(SELECT u.receiver, MAX(u.id) AS id 
      FROM tblA u 
      GROUP BY u.receiver 
    ) subset INNER JOIN tblA u ON subset.receiver = u.receiver AND subset.id = u.id, 
    (SELECT @ord := 0) init 
    ORDER BY points DESC 
) x 
ORDER BY rank; 
+----+--------+----------+----------+---------------------+--------+-------+------+ 
| id | sender | receiver | msg  | date    | points | refno | rank | 
+----+--------+----------+----------+---------------------+--------+-------+------+ 
| 9 | 2  | 4  | test ... | 2011-08-21 14:13:19 |  40 | 009 | 1 | 
| 3 | 4  | 2  | test ... | 2011-08-21 14:13:19 |  30 | 003 | 2 | 
| 5 | 1  | 3  | test ... | 2011-08-21 14:12:19 |  20 | 005 | 3 | 
+----+--------+----------+----------+---------------------+--------+-------+------+