棘手的SQL查詢

真的很棘手的SQL語句，我在這裏。嘗試構建這個查詢大約一小時。也許你可以幫助我。棘手的SQL查詢

我們有一張3列的表格： gamename |用戶| times_played

查詢應該選擇排名前三的遊戲（取決於總times_played）以及在該遊戲中玩過最多次數的前三名用戶=> 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

將是非常好的，如果你能幫助我=）謝謝！

來源

2009-10-09 samunai

什麼數據庫？ MySQL的？ SQL Server？甲骨文？ Postgres的？ – Asaph 2009-10-09 17:01:55

哪種技術？ MySQL的？ SQL Server？你可以使用存儲過程嗎？ – 2009-10-09 17:01:56

@foriamstu更好的問題：*你應該使用存儲過程嗎？根據我的經驗，幾乎總是：'不'。當時似乎很方便，回來咬你維修。 – 2009-10-10 03:39:45

更新：

由於@Steve Kass指出，我沒有注意到你只是想在前三場比賽。

下面是更新後的版本：

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

您將需要一個PRIMARY KEY這個查詢工作，假設它叫做id。

對此進行了更詳細的解釋在這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

來源

2009-10-09 16:59:44 Quassnoi

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

我想它是「PARTITION BY」 – 2009-10-09 17:01:19

我會在Game中發出命令並確保結果以正確的順序顯示。 – 2009-10-09 17:02:15

我不認爲Quassnoi注意到，您僅向前三名遊戲請求頂級用戶（根據總次數顯示）。這是一個查詢（沒有在真實數據上測試，因爲沒有給出CREATE TABLE和INSERT語句）。我還包括了Quassnoi沒有的關係，只是爲了向你展示這種選擇。

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;

來源

2009-10-10 03:36:21

謝謝你指出。有'12'upvotes和沒有做出，但一個simgle的話:) :) – Quassnoi 2009-10-10 19:23:54

也許你發佈是很少有錯，人們認爲這是浪費時間仔細觀察。 :) – 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

來源

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

來源

2013-07-10 09:09:17

檢查這一個 – 2013-07-10 09:09:39

棘手的SQL查詢

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