棘手的SQL查询

真的很棘手的SQL语句，我在这里。尝试构建这个查询大约一小时。也许你可以帮助我。棘手的SQL查询

我们有一张3列的表格： gamename |用户| times_played

查询应该选择排名前三的游戏（取决于总times_played）以及在该游戏中玩过最多次数的前三名用户=> 9行。

The result is like: 
CounterStrike | Smith 
CounterStrike | Jonny 
Counterstrike | Hans 
WoW   | George 
WoW   | Bob 
Wow   | Frank 
Need For Speed| James 
Need For Speed| Marion 
Need For Speed| Scarlet

将是非常好的，如果你能帮助我=）谢谢！

来源

2009-10-09 samunai

什么数据库？ MySQL的？ SQL Server？甲骨文？ Postgres的？ – Asaph 2009-10-09 17:01:55

哪种技术？ MySQL的？ SQL Server？你可以使用存储过程吗？ – 2009-10-09 17:01:56

@foriamstu更好的问题：*你应该使用存储过程吗？根据我的经验，几乎总是：'不'。当时似乎很方便，回来咬你维修。 – 2009-10-10 03:39:45

更新：

由于@Steve Kass指出，我没有注意到你只是想在前三场比赛。

下面是更新后的版本：

在SQL Server，Oracle和PostgreSQL 8.4：

SELECT gamename, user 
FROM (
     SELECT r.gamename, user, 
       ROW_NUMBER() OVER (PARTITION BY game ORDER BY times_played DESC) rn, 
     FROM (
       SELECT gamename, ROW_NUMBER() OVER (ORDER BY SUM(times_played) DESC) AS game_rn 
       FROM results 
       GROUP BY 
         gamename 
       ) g 
     JOIN results r 
     ON  r.gamename = g.gamename 
     WHERE game_rn <= 3 
     ) q 
WHERE rn <= 3 
ORDER BY 
     gamename, times_played DESC

在MySQL：

SELECT ro.gamename, ro.user 
FROM (
     SELECT gamename, SUM(times_played) AS rank 
     FROM results 
     ORDER BY 
       rank DESC 
     LIMIT 3 
     ) rd 
JOIN results ro 
ON  ro.gamename >= rd.gamename 
     AND ro.gamename <= rd.gamename 
     AND 
     (ro.times_played, ro.id) <= 
     (
     SELECT ri.times_played, ri.id 
     FROM results ri 
     WHERE ri.gamename = rd.gamename 
     ORDER BY 
       ri.times_played DESC, ri.id DESC 
     LIMIT 2, 1 
     ) 
ORDER BY 
     gamename, times_played DESC

您将需要一个PRIMARY KEY这个查询工作，假设它叫做id。

对此进行了更详细的解释在这篇文章中在我的博客：

Advanced row sampling

在PostgreSQL 8.3及以下：

SELECT gamename, ((ri)[s]).user 
FROM (
     SELECT gamename, ri, generate_series(1, 3) AS s 
     FROM (
       SELECT ro.gamename, 
         ARRAY 
         (
         SELECT ri 
         FROM results ri 
         WHERE ri.gamename = ro.gamename 
         ORDER BY 
           times_played DESC 
         LIMIT 3 
         ) AS ri 
       FROM (
         SELECT gamename, SUM(times_played) AS rank 
         FROM results 
         ORDER BY 
           rank DESC 
         LIMIT 3 
         ) rd 
       ) q 
     ) q2 
ORDER BY 
     gamename, s

来源

2009-10-09 16:59:44 Quassnoi

当你回答它时，它似乎很容易:-) – 2009-10-09 17:00:33

我想它是“PARTITION BY” – 2009-10-09 17:01:19

我会在Game中发出命令并确保结果以正确的顺序显示。 – 2009-10-09 17:02:15

我不认为Quassnoi注意到，您仅向前三名游戏请求顶级用户（根据总次数显示）。这是一个查询（没有在真实数据上测试，因为没有给出CREATE TABLE和INSERT语句）。我还包括了Quassnoi没有的关系，只是为了向你展示这种选择。

with GamesPlays(gamename,totalPlays) as (
    select 
    gamename, sum(times_played) 
    from results 
    group by gamename 
), GamesRanked(gamename,gameRank) as (
    select 
    gamename, 
    rank() over (
     order by totalPlays desc 
    ) 
    from GamesPlays 
), ResultsRanked(gamename,user,userRank) as (
    select 
    gamename, 
    user, 
    rank() over (
     partition by user 
     order by times_played desc 
    ) 
    from results; 
) 
    select 
    G.gamename, R.user 
    from ResultsRanked as R 
    join GamesRanked as G 
    on G.gamename = R.gamename 
    where gameRank <= 3 
    and userRank <= 3 
    order by 
    gameRank,userRank;

来源

2009-10-10 03:36:21

谢谢你指出。有'12'upvotes和没有做出，但一个simgle的话:) :) – Quassnoi 2009-10-10 19:23:54

也许你发布是很少有错，人们认为这是浪费时间仔细观察。 :) – 2009-10-11 01:04:10

DROP TABLE #game_stats 


CREATE TABLE #game_stats (gamename VARCHAR(50),users VARCHAR(50),times_played INT); 

INSERT INTO #game_stats VALUES ('Counter Strike','Kamesh',2); 
INSERT INTO #game_stats VALUES ('Counter Strike','Hely',4); 
INSERT INTO #game_stats VALUES ('Counter Strike','Maitri',1); 
INSERT INTO #game_stats VALUES ('Counter Strike','Laxmi',5); 
INSERT INTO #game_stats VALUES ('WOW','Kamesh',21); 
INSERT INTO #game_stats VALUES ('WOW','laxmi',60); 
INSERT INTO #game_stats VALUES ('WOW','Hely',7); 
INSERT INTO #game_stats VALUES ('NFS','Hely',5); 
INSERT INTO #game_stats VALUES ('NFS','Kamesh',1); 
INSERT INTO #game_stats VALUES ('NFS','Maitri',12); 
INSERT INTO #game_stats VALUES ('NFS','Laxmi',21); 
INSERT INTO #game_stats VALUES ('CODE ZERO','Kamesh',45); 
INSERT INTO #game_stats VALUES ('CODE ZERO','Maitri',52); 
INSERT INTO #game_stats VALUES ('CODE ZERO','Laxmi',21); 
INSERT INTO #game_stats VALUES ('CODE ZERO','Kamesh',41); 
INSERT INTO #game_stats VALUES ('HITMAN','Maitri',142); 
INSERT INTO #game_stats VALUES ('HITMAN','Laxmi',210); 
INSERT INTO #game_stats VALUES ('HITMAN','Kamesh',41); 
INSERT INTO #game_stats VALUES ('HITMAN','Maitri',102); 
INSERT INTO #game_stats VALUES ('HITMAN','Mani',142); 
INSERT INTO #game_stats VALUES ('NFS','Mani',210); 
INSERT INTO #game_stats VALUES ('CODE ZERO','Mani',41); 
INSERT INTO #game_stats VALUES ('WOW','Mani',102); 

select * from #game_stats; 

SELECT RN, 
Gamename, 
Users, 
Times_played 
FROM 
(
    SELECT ROW_NUMBER() OVER (PARTITION BY GS.gamename ORDER BY SUM(GS.times_played) DESC) AS RN, 
     GS.gamename, 
     GS.users, 
     SUM(gs.times_played) as times_played 

FROM #game_stats GS 
WHERE GS.gamename IN (
      SELECT TOP 3 gamename 
          FROM #game_stats 
          GROUP BY gamename 
                   ORDER BY sum(times_played) DESC 
        ) 
GROUP BY GS.gamename,GS.users 
) a 
WHERE RN<=3 
ORDER BY gamename,times_played DESC

来源

2011-11-03 06:06:56 Kamy

SELECT DISTINCT GN,US,GT,UT 
FROM 
(SELECT GN,US,GT,UT, 
     Dense_rank() over(ORDER BY GT DESC) RGT, 
     Dense_rank() over(partition BY GN ORDER BY UT DESC) RUT 
FROM 
    (SELECT gamename GN, 
      users US, 
      times_played TP, 
      sum(times_played) over (partition BY gamename) GT , 
      sum(times_played) over (partition BY gamename,users) UT 
    FROM game_stats)) 
WHERE RGT <4 
AND RUT < 4 
ORDER BY GT DESC, 
     UT DESC

来源

2013-07-10 09:09:17

检查这一个 – 2013-07-10 09:09:39

棘手的SQL查询

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