我有一个表格播放器的以下表格结构
Table Player {
Long playerID;
Long points;
Long rank;
}
假设 playerID 和 points 具有有效值，我可以根据单个查询中的点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，是不使用变量还是使用 HQL 来解决 hibernate 的限制？
【问题讨论】：
编辑：之前提出的更新语句不起作用。
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;
希望这会有所帮助。
【讨论】：
p1 引用，您将获得You can't specify target table 'p1' for update in FROM clause。
SELECT 时间工作，但它仍然无法处理关系。 OP 甚至将问题标记为“领带”！ :)
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;
【讨论】：
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)
如果你要求排名在平局的情况下跳过一个位置，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;
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。
【讨论】：
根据Normalization rules，排名应该在选择时进行评估。
【讨论】：