Question

对于表Player

，我有以下表结构

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

假设playerID和积分有有效值，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，他们应该为排名配合。

更新

我正在使用建议作为本机查询的查询来使用hibernate。 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？

Answer 1

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，而无需单独的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：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)

如果您要求等级跳过某个地方以防出现平局，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。

Answer 2

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;

Answer 3

编辑：前面提到的更新声明不起作用。

虽然这并不是您要求的：您可以在选择时动态生成排名：

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;

希望这有帮助。

Answer 4

根据Normalization rules，应在SELECT时评估等级。

更新MySQL表中的排名

4 个答案: