如何查询：agregate组由A列的每一列B值

Question

在这一刻，我有以下行的值：

构建此表：

CREATE TABLE `registros` (
    `id_registro` int(11) NOT NULL, 
    `id_partida` int(11) NOT NULL, 
    `id_juego` int(11) NOT NULL, 
    `id_jugador` int(11) NOT NULL, 
    `score` float NOT NULL DEFAULT '0', 
    `fecha_creacion` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; 

INSERT INTO `registros` (`id_registro`, `id_partida`, `id_juego`, `id_jugador`, `score`, `fecha_creacion`) VALUES 
(1, 2, 2, 1, 300, '2017-02-27 22:14:50'), 
(2, 2, 2, 2, 350, '2017-02-27 22:14:50'), 
(3, 2, 2, 3, 365, '2017-02-27 22:14:50'), 
(4, 2, 2, 4, 110, '2017-02-27 22:14:50'), 
(5, 2, 2, 5, 90, '2017-02-27 22:14:50'), 
(6, 2, 2, 6, 840, '2017-02-27 22:15:11'), 
(7, 2, 2, 7, 500, '2017-02-27 22:15:11'), 
(8, 2, 2, 1, 20, '2017-02-27 22:15:45'), 
(9, 2, 2, 1, 610, '2017-02-27 22:15:45'), 
(10, 2, 2, 2, 415, '2017-02-27 22:16:07'), 
(11, 2, 2, 4, 220, '2017-02-27 22:16:07'), 
(13, 3, 1, 1, -600, '2017-02-27 22:17:47'), 
(14, 3, 1, 1, -550, '2017-02-27 22:17:47'), 
(15, 3, 1, 2, -480, '2017-02-27 22:17:47'), 
(16, 3, 1, 2, -700, '2017-02-27 22:17:47'), 
(17, 3, 1, 9, -490, '2017-02-27 22:18:18'), 
(21, 3, 1, 2, -700, '2017-02-27 22:18:18'); 

我需要通过id_jugador与他的最好成绩为每id_partida（游戏大赛）（谁在比赛中发挥一定时期有不同的分数播放器），以得到一组。

我有一个“排行榜”每场比赛。这是我的SQL代码吧：

SELECT registros.id_partida, registros.id_jugador,MAX(registros.score) as best_score 
FROM registros 
WHERE registros.id_partida = 2 
GROUP BY registros.id_jugador 
ORDER BY best_score DESC; 

和执行结果：

但我想知道的所有比赛的所有级别和游戏中的位置。不仅使用where子句与特定目标id_partida = 2。究竟如下：

Mysql数据库在这里：http://pastebin.com/8eYuYzgV

谢谢大家。

Answer 1

您可以按两列组：

SELECT registros.id_partida, registros.id_jugador,MAX(registros.score) as best_score 
FROM registros 
GROUP BY registros.id_jugador, registros.id_partida 
ORDER BY best_score DESC; 

如果你想在你的查询中的排名则查询会显得稍微复杂与MySQL：

select id_partida, id_jugador, best_score, rank 
FROM (
select *, 
case when @p=a.id_partida THEN @o:[email protected]+1 ELSE @o:=1 END as rank, 
@p:=a.id_partida 
from (
SELECT registros.id_partida, registros.id_jugador,MAX(registros.score) as best_score 
FROM registros 
GROUP BY registros.id_jugador, registros.id_partida 
ORDER BY registros.id_partida, best_score DESC 
) a, (select @p:=0, @o:=1) s 
) scores 

结果：

| id_partida | id_jugador | best_score | rank | 
|------------|------------|------------|------| 
|   2 |   6 |  840 | 1 | 
|   2 |   1 |  610 | 2 | 
|   2 |   7 |  500 | 3 | 
|   2 |   2 |  415 | 4 | 
|   2 |   3 |  365 | 5 | 
|   2 |   4 |  220 | 6 | 
|   2 |   5 |   90 | 7 | 
|   3 |   2 |  -480 | 1 | 
|   3 |   9 |  -490 | 2 | 
|   3 |   1 |  -550 | 3 | 

SQL Fiddle