Question

用单词来解释我试图完成的事情是非常困难的。所以我试着做一个反映我真实项目的例子。我有一群人（Peter，Aaron，Mark，Alicia，Cleo和Mike）互相玩游戏（A，B和C）。我想用我的查询做的是让所有组合对于特定游戏互相对战。例如：

彼得打了一场比赛，对阵亚伦，他在3轮比赛中获胜 - ＆gt;彼得 - 亚伦 - A - 1 - 5 - 1
迈克与克莱奥队进行了两场C场比赛，并且在3轮和4轮比赛中两次失利 - > 迈克 - 克莱奥 - C - 0 - 3.5 - 2

桌面游戏示例：

id   playRound   player   game   win   rounds 

1    1           Peter    A      1     5
2    1           Aaron    A      0     5
3    1           Alicia   B      0     3 
4    1           Mark     B      1     3
5    1           Mike     C      0     4
6    1           Cleo     C      1     4
7    2           Alicia   A      1     5
8    2           Mark     A      0     5
9    2           Peter    B      1     3
10   2           Mark     B      0     3
11   2           Mike     C      0     3
12   2           Cleo     C      1     3
13   3           Alicia   A      0     4
14   3           Mark     A      1     4
15   3           Peter    B      0     4
16   3           Aaron    B      1     4
17   3           Alicia   C      1     5
18   3           Cleo     C      0     5

最终结果：

playerONE   playerTwo   game   avgWinRateOne   avgRounds   Number

Peter       Aaron       A      1               5           1
Peter       Aaron       B      1               3           1
Peter       Mark        B      0               4           1
Alicia      Mark        A      0.5             4.5         2
Alicia      Mark        B      0               3           1
Alicia      Cleo        C      1               5           1
Mike        Cleo        C      0               3.5         2

我正在摆弄这个，但我不知道我在做什么

SELECT *
FROM (

SELECT *
FROM `testtable`
WHERE `game` = 'A'
GROUP BY `gameId` , `game` , `player`
)tmp1, (

SELECT *
FROM `testtable`
WHERE `game` = 'A'
GROUP BY `gameId` , `game` , `player`
)tmp2

Answer 1

我想也许你想要这个查询：

select p1.player, p2.player, p1.game, avg(p1.win), avg(p1.rounds), count(*)
  from games p1
    inner join games p2
      on p1.playround = p2.playround
        and p1.game = p2.game
        and p1.player != p2.player
  group by p1.player, p2.player, p1.game;

它是一个相对基本的join，我们将游戏表与自身相关联，以获得我们的游戏配对，我们可以group by这些配对，以便使用聚合函数来计算平均值和比赛次数。

在样本数据上运行它会得到以下结果：

mysql> select p1.player, p2.player, p1.game, avg(p1.win), avg(p1.rounds), count(*)
    ->   from games p1
    ->     inner join games p2
    ->       on p1.playround = p2.playround
    ->         and p1.game = p2.game
    ->         and p1.player != p2.player
    ->   group by p1.player, p2.player, p1.game;
+--------+--------+------+-------------+----------------+----------+
| player | player | game | avg(p1.win) | avg(p1.rounds) | count(*) |
+--------+--------+------+-------------+----------------+----------+
| Aaron  | Peter  | A    |      0.0000 |         5.0000 |        1 |
| Aaron  | Peter  | B    |      1.0000 |         4.0000 |        1 |
| Alicia | Cleo   | C    |      1.0000 |         5.0000 |        1 |
| Alicia | Mark   | A    |      0.5000 |         4.5000 |        2 |
| Alicia | Mark   | B    |      0.0000 |         3.0000 |        1 |
| Cleo   | Alicia | C    |      0.0000 |         5.0000 |        1 |
| Cleo   | Mike   | C    |      1.0000 |         3.5000 |        2 |
| Mark   | Alicia | A    |      0.5000 |         4.5000 |        2 |
| Mark   | Alicia | B    |      1.0000 |         3.0000 |        1 |
| Mark   | Peter  | B    |      0.0000 |         3.0000 |        1 |
| Mike   | Cleo   | C    |      0.0000 |         3.5000 |        2 |
| Peter  | Aaron  | A    |      1.0000 |         5.0000 |        1 |
| Peter  | Aaron  | B    |      0.0000 |         4.0000 |        1 |
| Peter  | Mark   | B    |      1.0000 |         3.0000 |        1 |
+--------+--------+------+-------------+----------------+----------+
14 rows in set (0.00 sec)

选择所有组合

1 个答案: