查询方法：对B列的每一列值按A列聚合分组

Question

此时我有以下行值：

构建此 table 与：

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');

而且我需要 id_jugador 为每个 id_partida （游戏比赛）获得他最好的分数（一个玩家有时以不同的分数玩游戏）。

每场比赛我都有一个"ranking"。这是我的 sql 代码：

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;

执行结果：

但是我想知道所有比赛的所有排名和比赛的位置。不仅使用带有特定目标的 where 子句 id_partida = 2。具体如下：

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

谢谢大家

Answer 1

您可以按两列分组：

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;

如果您希望在查询中获得排名，那么查询看起来会稍微复杂 MySQL:

select id_partida, id_jugador, best_score, rank
FROM (
select *,
case when @p=a.id_partida THEN @o:=@o+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

查询方法：对B列的每一列值按A列聚合分组

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innodb