8

我有一个表播放器的以下表结构

Table Player {  
Long playerID;  
Long points;  
Long rank;  
}

假设 playerID 和积分具有有效值,我可以根据单个查询中的积分数更新所有玩家的排名吗?如果两个人的分数相同,他们应该并列排名。

更新:

我正在使用建议作为本机查询的查询来使用休眠。Hibernate 不喜欢使用变量,尤其是':'。有谁知道任何解决方法?在这种情况下,通过不使用变量或通过使用 HQL 来解决休眠的限制?

4

4 回答 4

17

一种选择是使用排名变量,例如:

UPDATE   player
JOIN     (SELECT    p.playerID,
                    @curRank := @curRank + 1 AS rank
          FROM      player p
          JOIN      (SELECT @curRank := 0) r
          ORDER BY  p.points DESC
         ) ranks ON (ranks.playerID = player.playerID)
SET      player.rank = ranks.rank;

JOIN (SELECT @curRank := 0)部分允许变量初始化而不需要单独的SET命令。

关于这个主题的进一步阅读:


测试用例:

CREATE TABLE player (
   playerID int,
   points int,
   rank int
);

INSERT INTO player VALUES (1, 150, NULL);
INSERT INTO player VALUES (2, 100, NULL);
INSERT INTO player VALUES (3, 250, NULL);
INSERT INTO player VALUES (4, 200, NULL);
INSERT INTO player VALUES (5, 175, NULL);

UPDATE   player
JOIN     (SELECT    p.playerID,
                    @curRank := @curRank + 1 AS rank
          FROM      player p
          JOIN      (SELECT @curRank := 0) r
          ORDER BY  p.points DESC
         ) ranks ON (ranks.playerID = player.playerID)
SET      player.rank = ranks.rank;

结果:

SELECT * FROM player ORDER BY rank;

+----------+--------+------+
| playerID | points | rank |
+----------+--------+------+
|        3 |    250 |    1 |
|        4 |    200 |    2 |
|        5 |    175 |    3 |
|        1 |    150 |    4 |
|        2 |    100 |    5 |
+----------+--------+------+
5 rows in set (0.00 sec)

更新:刚刚注意到您需要领带才能共享相同的排名。这有点棘手,但可以通过更多变量来解决:

UPDATE   player
JOIN     (SELECT    p.playerID,
                    IF(@lastPoint <> p.points, 
                       @curRank := @curRank + 1, 
                       @curRank)  AS rank,
                    @lastPoint := p.points
          FROM      player p
          JOIN      (SELECT @curRank := 0, @lastPoint := 0) r
          ORDER BY  p.points DESC
         ) ranks ON (ranks.playerID = player.playerID)
SET      player.rank = ranks.rank;

对于一个测试用例,让我们添加另一个 175 分的玩家:

INSERT INTO player VALUES (6, 175, NULL);

结果:

SELECT * FROM player ORDER BY rank;

+----------+--------+------+
| playerID | points | rank |
+----------+--------+------+
|        3 |    250 |    1 |
|        4 |    200 |    2 |
|        5 |    175 |    3 |
|        6 |    175 |    3 |
|        1 |    150 |    4 |
|        2 |    100 |    5 |
+----------+--------+------+
6 rows in set (0.00 sec)

如果您要求排名在平局的情况下跳过一个位置,您可以添加另一个IF条件:

UPDATE   player
JOIN     (SELECT    p.playerID,
                    IF(@lastPoint <> p.points, 
                       @curRank := @curRank + 1, 
                       @curRank)  AS rank,
                    IF(@lastPoint = p.points, 
                       @curRank := @curRank + 1, 
                       @curRank),
                    @lastPoint := p.points
          FROM      player p
          JOIN      (SELECT @curRank := 0, @lastPoint := 0) r
          ORDER BY  p.points DESC
         ) ranks ON (ranks.playerID = player.playerID)
SET      player.rank = ranks.rank;

结果:

SELECT * FROM player ORDER BY rank;

+----------+--------+------+
| playerID | points | rank |
+----------+--------+------+
|        3 |    250 |    1 |
|        4 |    200 |    2 |
|        5 |    175 |    3 |
|        6 |    175 |    3 |
|        1 |    150 |    5 |
|        2 |    100 |    6 |
+----------+--------+------+
6 rows in set (0.00 sec)

注意:请考虑我建议的查询可以进一步简化。

于 2010-04-28T06:25:15.557 回答
6

丹尼尔,你有很好的解决方案。除了一点 - 领带盒。如果 3 名玩家之间出现平局,此更新将无法正常工作。我将您的解决方案更改如下:

UPDATE player  
    JOIN (SELECT p.playerID,  
                 IF(@lastPoint <> p.points,  
                    @curRank := @curRank + @nextrank,  
                    @curRank)  AS rank,  
                 IF(@lastPoint = p.points,  
                    @nextrank := @nextrank + 1,  
                    @nextrank := 1),  
                 @lastPoint := p.points  
            FROM player p  
            JOIN (SELECT @curRank := 0, @lastPoint := 0, @nextrank := 1) r  
           ORDER BY  p.points DESC  
          ) ranks ON (ranks.playerID = player.playerID)  
SET player.rank = ranks.rank;
于 2011-10-22T14:42:23.537 回答
3

编辑:之前提出的更新声明不起作用。

虽然这不是您所要求的:您可以在选择时即时生成排名:

select p1.playerID, p1.points, (1 + (
    select count(playerID) 
      from Player p2 
     where p2.points > p1.points
    )) as rank
from Player p1
order by points desc

编辑:再次尝试 UPDATE 语句。临时表怎么样:

create temporary table PlayerRank
    as select p1.playerID, (1 + (select count(playerID) 
                                   from Player p2 
                                  where p2.points > p1.points
              )) as rank
         from Player p1;

update Player p set rank = (select rank from PlayerRank r 
                             where r.playerID = p.playerID);

drop table PlayerRank;

希望这可以帮助。

于 2010-04-28T06:06:32.187 回答
0

根据规范化规则,排名应该在 SELECT 时进行评估。

于 2010-04-28T06:08:38.240 回答