更新MySQL表中的排名

sma*_*esh 8 mysql hibernate tie rank

我有一个表Player的以下表结构

Table Player {  
Long playerID;  
Long points;  
Long rank;  
}
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假设玩家ID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等级吗?如果两个人拥有相同数量的积分,那么他们应该排名等级.

更新:

我正在使用hibernate使用建议作为本机查询的查询.Hibernate不喜欢使用变量,特别是':'.有谁知道任何变通方法?通过不使用变量或在这种情况下使用HQL解决hibernate的限制?

Dan*_*llo 16

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;
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JOIN (SELECT @curRank := 0)部分允许变量初始化而无需单独的SET命令.

进一步阅读这个主题:


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;
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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)
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更新:刚刚注意到您需要联系以共享相同的排名.这有点棘手,但可以通过更多变量解决:

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;
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对于测试用例,让我们添加另一个175分的玩家:

INSERT INTO player VALUES (6, 175, NULL);
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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)
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如果你要求排名跳过一个地方以防平局,你可以添加另一个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;
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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)
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注意:请考虑我建议的查询可以进一步简化.


小智 6

丹尼尔,你有非常好的解决方案.除了一点 - 领带案例.如果3个玩家之间发生平局,则此更新无效.我改变了你的解决方案如下:

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;
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