1

我的架构

我有以下表格

table             notes/example values
------------------------------------------------
users (
  id
  email           # "foo@example.com"
)                 

games (           
  id              
  name            # "Space Invaders", "Asteroids", "Centipede"
)                 

players (         
  id              
  name            # "uber dude"
  user_id         # player belongs to user
  game_id         # player belongs to game
)                 

scores (          
  id              
  player_id       # belongs to one player
  value           # 50
  created_at      # "2010-09-10",   "2010-08-05"
  month           # "2010-09",      "2010-08"
)

我需要创建两个报告。

1) 顶级球员

最近 4 个月表现最好的球员（每位球员的所有得分之和）。显示每个月的前 10 名。

    2010-07         2010-08           2010-09    2010-10
 1  plyA 5,000 pts  plyB  9,400 pts   ...        ...
    Centipede       Solitaire

 2  plyB 3,600 pts  plyC  8,200 pts   ...        ...
    Asteroids       Centipede       

 3  plyC 2,900 pts  plyA  7,000 pts   ...        ...
    Centipede       Centipede

 4  ...             ...               ...        ...
 5  ...             ...               ...        ...
 6  ...             ...               ...        ...
 7  ...             ...               ...        ...
 8  ...             ...               ...        ...
 9  ...             ...               ...        ...
10  ...             ...               ...        ...

2) 热门用户：

最近 4 个月内表现最好的用户（每个用户的每个玩家的所有得分总和）。显示每个月的前 10 名。

    2010-07           2010-08             2010-09    2010-10
 1  userA 50,000 pts  userB 51,400 pts    ...        ...
 2  userB 40,500 pts  userA 39,300 pts    ...        ...
 3  userC 40,200 pts  userC 37,000 pts    ...        ...
 4  ...               ...                 ...        ...
 5  ...               ...                 ...        ...
 6  ...               ...                 ...        ...
 7  ...               ...                 ...        ...
 8  ...               ...                 ...        ...
 9  ...               ...                 ...        ...
10  ...               ...                 ...        ...

MySQL 视图助手

出于加入目的，我有一个存储视图来帮助查询报告的月份。它将始终返回最近的 4 个月。

report_months (
  month
)

SELECT * FROM report_months;

2010-07
2010-08
2010-09
2010-10

问题

例如，在报告 #1 中，我可以很容易地得到总和。

select
  p.name        as player_name,
  g.name        as game_name,
  s.month       as month,
  sum(s.score)  as sum_score

from players  as p

join games    as g
  on g.id = p.game_id

join scores   as s
  on s.player_id = p.id

join report_months as rm  -- handy view helper
  on rm.month = s.month

group by
  p.name, g.name

order by
  sum(s.score) desc

-- I can't do this :(
-- limit 0, 40

但是，我不能简单地获取前 40 个结果并将它们分散到 4 个月，因为这不能保证我每个月都有 10 个。

问题

如何修改我的查询以确保每个月获得 10 个？

4

1 回答 1

2

正如您所展示的，我不会尝试进行按月制表的 SQL 查询。

相反，以行而不是列的形式查询每月前 10 名玩家：

Month    Rank  Player  TotalScore  Game
2010-07     1    plyA   5,000 pts  Centipede
2010-07     2    plyB   3,600 pts  Asteroids
2010-07     3    plyC   2,900 pts  Centipede
...
2010-08     1    plyB   9,400 pts  Solitaire
2010-08     2    plyC   8,200 pts  Centipede
2010-08     3    plyA   7,000 pts  Centipede
...

这成为一个greatest-n-per-group问题，n10在哪里。 

CREATE VIEW PlayerScoresByMonth AS
  SELECT month, player_id, SUM(value) AS score
  FROM scores
  GROUP BY month, player_id;

SELECT s1.month, COUNT(s2.month)+1 AS Rank, s1.player_id, s1.score AS TotalScore
FROM PlayerScoresByMonth s1
LEFT OUTER JOIN PlayerScoresByMonth s2 ON s1.month = s2.month 
  AND (s1.score < s2.score OR s1.score = s2.score AND s1.player_id < s2.player_id)
GROUP BY s1.month, s1.player_id
HAVING COUNT(*) < 10
ORDER BY s1.month, Rank;

（这是未经测试的，但应该让你开始）

然后，您需要编写一些应用程序代码来获取此查询的结果并按月份分隔列表，并呈现您打算这样做的数据。

于 2010-10-20T20:20:39.807 回答