問題與SQL SELECT查詢

我有此信息的桌牌遊戲：問題與SQL SELECT查詢

- Id_Game (int) 
- Id_Referee1 (int) 
- Id_Referee2 (int)

這些都是一些行：

ID_GAME ID_REFEREE1 ID_REFEREE2 
1   1000   728 
2   1004   813 
3   728   1004 
4   1004   1000

我想和遊戲，一個數的列表裁判已主持，無論作爲裁判1還是裁判2都無所謂。在這種情況下，我希望通過DESC遊戲的數量收到此列表順序。

REFEREE NUMBER OF GAMES 
1004  3 
728  2 
1000  2 
813  1

這將是SQL查詢？我不知道如何連接字段Id_Referee1和Id_Referee2 ...

THANKS

來源

2010-12-13 Arturo

我想：

select referee, count(*) from (
    select id_game, id_referee1 as referee 
    from games 
    union 
    select id_game, id_referee2 as referee 
    from games 
) 
group by referee

來源

2010-12-13 11:31:03

噢，你是更快:) – 2010-12-13 11:33:27

你假設id_game是獨一無二的 - 當然這可能是，但@Arturo沒有指定 – 2010-12-13 11:53:33

@JackPDouglas你說得對，但你假設裁判1！=裁判2 - 哪他也沒有指定;） – 2010-12-13 11:56:59

假設id_referee1 = id_referee2所有可能的行（可能使用檢查約束）：

select referee, count(*) as number_of_games 
from(select id_referee1 as referee from games 
     union all 
     select id_referee2 from games) 
group by referee 
order by count(*) desc;

，如果你不想做這樣的假設：

select referee, count(*) as number_of_games 
from(select id_referee1 as referee from games 
     union all 
     select id_referee2 
     from games 
     where id_referee1 is null or id_referee1!=id_referee2) 
group by referee 
order by count(*) desc;

來源

2010-12-13 11:52:37

SELECT ID_REFEREE1, count(*) as no_of_games 
FROM games 
GROUP BY ID_REFEREE1 
ORDER BY 2

來源

2010-12-13 12:04:34

如果您發佈的是代碼或XML，**請**在文本編輯器中突出顯示這些行，然後單擊編輯器工具欄上的「代碼」按鈕（101 010），以更好地格式化和語法突出顯示它！ – 2010-12-13 12:19:49

用於和平格式化 – 2010-12-13 12:25:35

我認爲某件事情是錯誤的，因爲我得到一個錯誤與此查詢：

SELECT Referee, count(*) FROM 
(
    (SELECT Id_Game, Id_Referee1 as Referee FROM Games) 
    UNION ALL 
    (SELECT Id_Game, Id_Referee2 as Referee FROM Games) 
) GROUP BY Referee

你可以看到什麼是錯的？感謝

來源

2010-12-13 13:33:45 Arturo

事實上，我得到的錯誤是「＃1248 - 每個派生表都必須有它自己的別名」。任何幫助？ – Arturo 2010-12-13 13:54:35

有些RDBMS（例如MySQL，postgres）堅持爲每個子查詢設置一個別名，因此''select * from（select * from games）'拋出一個錯誤，但是'select * from（select * from game）'因爲g'很好。 – 2010-12-13 14:06:17

問題與SQL SELECT查詢

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