在 PHP/MySQL 中计算排名

Question

我在 MySQL 中有一个表，比如说它有两个字段 Username、GameName 和 Score。 我想计算个人游戏名称的用户排名，以便进行查询

SELECT * FROM scores WHERE `GameName` = 'Snake' ORDER BY `Score` DESC

按从高到低的顺序获取所有用户的列表，并为每个用户分配一个编号。

但是有没有更简单的方法来获得单个用户的排名，而不是选择整个表格，因为这似乎不太有效。

谢谢

Answer 1

如果您想要整体排名，很遗憾您必须对整个表格进行排序。简而言之，如果不知道表中的其他排名，就无法知道某人在表中的排名。

也就是说，如果您担心性能，这里有一个相当简单的解决方案 - 缓存您的排名查询结果（可能到另一个 MySQL 表中！），并查询所有读取。当有人发布新分数时，重新计算您的临时表。您可以定期刷新某个排名以下的所有记录（例如，排名低于 100 的任何人都会从分数表中删除）以保持快速重新计算，因为没有人会在被更高的分数击倒后提升排名。

# Create your overall leaderboards once
create table leaderboards (rank integer primary key, score_id integer, game varchar(65), user_id integer, index game_user_id_idx (game, user_id))


# To refresh your leaderboard, we'll query the ranks for the game into a temporary table, flush old records from scores, then copy
# the new ranked table into your leaderboards table.
# We'll use MySQL's CREATE TABLE...SELECT syntax to select our resultset into it directly upon creation.
create temporary table tmp_leaderboard (rank integer primary key auto_increment, score_id integer, game varchar(65), user_id integer)
  select ID, GameName, UserID, from scores where GameName = '$game' order by score desc;

# Remove old rankings from the overall leaderboards, then copy the results of the temp table into it.
delete from leaderboards where game = '$game';
insert into leaderboards (rank, score_id, game, user_id)
  select rank, score_id, game, user_id from tmp_leaderboard;

# And then clean up the lower scores from the Scores table
delete from scores join tmp_leaderboard on scores.id = tmp_leaderboard.score_id, scores.GameName = tmp_leaderboard.game where tmp_leaderboard.rank < 100;

# And we're done with our temp table
drop table tmp_leaderboard;

然后，每当您想读取游戏的排名时：

select rank from leaderboards where game = '$game' and user_id = '$user_id';

Answer 2

您无法避免读取表中的大量数据 - 但您无需将其一直拖回处理脚本：

SELECT COUNT(*)
FROM scores 
WHERE `GameName` = 'Snake'
AND user=$some_user;

（由于您可能希望第一个人的排名为“1”而不是“0”，因此增加结果）。

但是，如果您需要经常运行查询，则值得维护排序结果的具体化视图。

Answer 3

从您的用户表中获取用户 ID 并在您的查询中使用它

SELECT * FROM scores WHERE `GameName` = 'Snake' 
and `youruseridfield` = '$useridvalue'
ORDER BY `Score` DESC

Answer 4

SELECT * FROM scores WHERE 'GameName' = 'Snake' && userID = '$userID' ORDER BY 'Score' DESC

Answer 5

看看有没有办法在 MySQL 中获得排名，但这里是你可以在 PHP 中做到的方法：

function getRank($user, $game, $limit=50) {
    $sql = "
SELECT @rank:=@rank+1 AS Rank, User, GameName
FROM scores, (SELECT @rank:=1) AS i
WHERE `GameName` = '$game' 
ORDER BY `Score` DESC
LIMIT 0, $limit
";

    $result =  mysql_query($sql);

    while ($row = mysql_fetch_assoc($result)) {
        if ($row['User'] == $user) {
            return $row['Rank'];
        }
    }

    return -1;
}

注意，我把限制放在那里，否则你只会得到 30 个结果。如果玩家未排名，则返回 -1。

Answer 6

我的一个查询解决方案：

select @rank:=@rank+1 AS Rank,L1.* from 
    (select @rank:=0) as L2,
    (select i.*,count(*) as sum 
        FROM 
        transactions t
        LEFT JOIN companies c on (c.id = t.company_id)  
        LEFT JOIN company_industries ci on (c.id = ci.company_id)  
        LEFT JOIN industries i on (ci.industry_id = i.id)
        GROUP by i.name 
        ORDER BY sum desc ) as L1;

Answer 7

您应该为 GameName 和 Score 添加索引。 （并且不要忘记在插入查询之前转义 GameName => mysql_real_escape_string）。 @Bryan：不应该是“AND”而不是“&&”吗？

Answer 8

下面的查询使用了rank函数，这也可以实现：

WITH game_rank AS(
    SELECT 
        user_id, 
        gameName, 
        RANK() OVER (
            PARTITION BY gameName
            ORDER BY score DESC
        ) order_value_rank
    FROM
        scores
)
SELECT 
    * 
FROM 
    game_rank
WHERE 
    gameName = "Snake";