我正在尝试获得最高分行以及排名并选择没有 rownum=rownum+1 的查询。我已经尝试了下面的查询，但我还遗漏了一些链接 http://sqlfiddle.com/#!2/fd7897/7 .

我正在寻找答案，例如每个接收者的最后一个条目，这也是排名最高的条目:我非常感谢任何帮助。提前致谢。

像这样:

('2', '4', 'test ...', '2011-08-21 14:13:19', '40', '009')---rank1

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

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

目前查询:

    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;

表格:

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');

最佳答案

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:

要将其创建为 MySQL View ，您必须正确处理 from 子句中的聚合:

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;

关于mysql - 按最大 ID 和点数对用户进行排名，但(排名错误)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21715642/