與分組的複雜的總和

我有表名爲遊戲與列：Player1Id，Player2Id，Player1Points，Player2Points。與分組的複雜的總和

的樣本數據：

Player1Id Player2Id Player1Points Player2Points 
--------- --------- ------------- ------------- 
John   Piter  4     1 
John   Adam   2    10 
Piter  Adam   4     2

而且我想與點的總和列表每個球員，這樣的：

PlayerId Points 
-------- ------ 
John   6 
Piter  5 
Adam  12

如何才達到，在SQL（SQL Server 2008中）？如果只有某位球員贏得比賽，我需要總積分？它可以不使用變量，循環等？

來源

2010-06-22 Abc

SELECT Player1Id AS PlayerId, SUM(Player1Points) AS Points FROM 
(SELECT Player1Id, Player1Points FROM MyTable 
UNION ALL 
SELECT Player2Id, Player2Points FROM MyTable) t1 
GROUP BY Player1Id 
ORDER BY Player1Id

產量：

PlayerId  Points 
--------  ------ 
Adam    12 
John    6 
Piter    5

如果你添加更多的玩家查詢將獲得更加複雜。最好在不同的行上記錄每個玩家的分數。你可以添加一列來指定遊戲。

PlayerId Points GameId 
-------- ------ ------ 
John   4  1 
Piter   1  1 
John   2  2 
Adam   10  2 
Piter   4  3 
Adam   2  3

您的查詢將被簡單：

SELECT PlayerId, SUM(Points) 
FROM MyTable 
GROUP BY PlayerId 
ORDER BY PlayerId

來源

2010-06-22 14:28:14

這應該是'UNION ALL'（如果約翰在兩場比賽中得分3，它只會計算一次） – Quassnoi 2010-06-22 14:31:19

@Quassnoi，修正了，謝謝！ – 2010-06-22 14:33:27

我會忍不住重新組織數據，也許是：

Gameid Player Score 
    1 John  4 
    2 John  2 
    1 Piter 1

等

來源

2010-06-22 14:30:17 Patrick

另外，如果您開始與3名玩家進行遊戲，則不必添加列。 + 1用於建議標準化數據。 – HLGEM 2010-06-22 14:42:09

測試表：

DECLARE @Table TABLE (Player1Id VARCHAR(17), Player2Id VARCHAR(17), Player1Points INT, Player2Points INT); 
INSERT INTO @Table 
    SELECT 'John', 'Piter', 4, 1 UNION ALL 
    SELECT 'John', 'Adam', 2, 10 UNION ALL 
    SELECT 'Piter', 'Adam', 4, 2;

個

總積分：從贏得比賽

WITH CTE (PlayerId, Points) AS (
    SELECT Player1Id, Player1Points FROM @Table 
    UNION ALL 
    SELECT Player2Id, Player2Points FROM @Table 
) 
SELECT PlayerId, SUM(Points) FROM CTE GROUP BY PlayerId;

總積分：

WITH CTE (PlayerId, Points, Won) AS (
    SELECT Player1Id, Player1Points, CASE WHEN Player1Points > Player2Points THEN 1 ELSE 0 END FROM @Table 
    UNION ALL 
    SELECT Player2Id, Player2Points, CASE WHEN Player2Points > Player1Points THEN 1 ELSE 0 END FROM @Table 
) 
SELECT PlayerId, SUM(Points) FROM CTE WHERE Won = 1 GROUP BY PlayerId;

差異：

WITH CTE (PlayerId, Points) AS (
    SELECT Player1Id, CASE WHEN Player1Points > Player2Points THEN Player1Points - Player2Points ELSE (Player2Points - Player1Points) * -1 END FROM @Table 
    UNION ALL 
    SELECT Player2Id, CASE WHEN Player2Points > Player1Points THEN Player2Points - Player1Points ELSE (Player1Points - Player2Points) * -1 END FROM @Table 
) 
SELECT PlayerId, SUM(Points) FROM CTE GROUP BY PlayerId;

來源

2010-06-22 14:32:53 Don

這隻會計算分數的勝利：

SELECT player, SUM(score) AS score 
FROM (
     SELECT player1id AS player, player1points AS score 
     FROM mytable 
     WHERE player1points > player2points 
     UNION ALL 
     SELECT player2id, player2points 
     FROM mytable 
     WHERE player2points > player1points 
     ) q 
GROUP BY 
     player

來源

2010-06-22 14:33:23 Quassnoi

要回答你的第二個問題，然後你可以改變馬庫斯的答案是：

SELECT 
    Player1Id, 
    SUM(Player1Points) 
FROM 
(
    SELECT Player1Id, Player1Points FROM MyTable WHERE Player1Points >= Player2Points 
    UNION ALL 
    SELECT Player2Id, Player2Points FROM MyTable WHERE Player2Points >= Player1Points 
) t1 
GROUP BY 
    Player1Id

根據您想如何處理的關係，你需要改變> =。

來源

2010-06-22 14:35:58

與分組的複雜的總和

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