Question

此时我有以下行值：

使用以下代码构建此表：

node_modules

我需要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');以id_jugador（游戏比赛）的最佳分数（一个在游戏中玩不同分数的玩家）得到一个小组。{/ p>

我有一个＆＃34;排名＆＃34;对于每场比赛。这是我的sql代码：

id_partida

执行结果：

但我想知道所有比赛的所有级别和比赛的位置。不仅要将where子句与特定目标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;一起使用。完全如下：

这里的Mysql数据库：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;

如果您想在查询中使用排名，那么查询看起来会与MySQL略微复杂：

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

如何查询：按列A为每个列B值聚集组

1 个答案: