在MySQL表中更新等级

StackOverflow https://stackoverflow.com/questions/2727138

  •  02-10-2019
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题

我有一个表播放器的下表结构 

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

假设播放器和点具有有效值，我可以根据单个查询中的点数更新所有播放器的排名吗？如果两个人的分数相同，那么他们应该领先。

更新：

我正在使用建议的查询作为本机查询。 Hibernate不喜欢使用变量，尤其是“：”。有人知道有任何解决方法吗？在这种情况下，不使用变量或在这种情况下使用HQL来解决Hibernate的限制？

有帮助吗？

解决方案

一种选择是使用排名变量，例如以下内容：

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;

这 JOIN (SELECT @curRank := 0) 零件允许可变初始化，而无需单独 SET 命令。

有关此主题的进一步阅读：

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： 只是注意到您需要纽带才能共享相同的等级。这有点棘手，但是可以用更多的变量来解决：

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，让我们添加另一个以175分的球员：

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)

而且，如果您需要等级在领带时跳过一个地方，则可以添加另一个 IF 健康）状况：

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。

其他提示

丹尼尔，您有很好的解决方案。除了一个点 - 领带盒。如果在3个播放器之间进行平局，此更新无法正常工作。我更改了您的解决方案如下：

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;

编辑：前面介绍的更新语句无效。

尽管这并不是您要要求的：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：再次尝试更新语句。临时桌子怎么样：

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。

根据 归一化规则, ，应在选定时间评估等级。

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