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在存储锦标赛的数据库中，我有这种关系结构来存储比赛及其比赛：

EER图

创建比赛时，根据比赛中的比赛数量插入一个比赛行和多个游戏行。然后还插入 Match_left 和 match_right 球员行，以将比赛中的两名球员与比赛行相关联。当玩家被记录为赢得游戏时，将与相应游戏相关的行插入到 game_winners 表中。有没有办法编写一个查询来从赢得比赛中得出谁赢得了比赛？也许从逻辑上讲，这应该是一个视图，以便我可以随时轻松地将匹配表行加入它？

这是一些示例数据

我希望 Fred 作为 match_ID 1 的获胜者返回，因为他赢得了 2/3 场比赛。我希望 match_ID 2 的获胜者为空，因为两名球员都没有赢得必要的比赛数（2/3）。

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2 回答 2

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我在这里假设一场比赛的获胜者是赢得最多比赛的球员。

对比赛中的球员进行排名很简单。

SELECT player_id, COUNT(*) AS game_count
  FROM game_winners
  WHERE match_id = ?
  GROUP BY player_id
  ORDER BY COUNT(*) DESC

一次找到不止一场比赛的获胜者有点棘手。

SELECT match_id, player_id, COUNT(*) AS game_count
  FROM game_winners
  GROUP BY match_id, player_id

让我们调用上面的 player_wins

SELECT match_id, MAX(game_count) AS match_winner
  FROM player_wins
  GROUP BY match_id

这将为您提供每场比赛的获胜分数。

SELECT match_id, player_id
  FROM player_wins
  INNER JOIN (
    SEELCT match_id, MAX(game_count) AS match_winner
    FROM player_wins
    GROUP BY match_id
  ) AS winning_score
  ON player_wins.match_id = winning_score.match_id
  AND player_wins.game_count = winning_score.match_winner

这为您提供了每场比赛的获胜球员。

于 2011-08-05T21:22:09.010 回答
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让我们看看我是否正确理解了您的示例数据。

SELECT match_id, round_id, COUNT(*) AS game_count
  FROM games
  GROUP BY match_id, round_id

这为您提供了每场比赛的比赛次数。比赛是最好的赛制，所以你必须赢得超过一半的比赛场数才能赢得比赛。

SELECT match_id, round_id, player_id, COUNT(*) AS win_count
  FROM game_winners
  GROUP BY match_id, round_id

这为您提供了每场比赛中每位球员赢得的比赛数。

SELECT match_id, round_id, player_id
  FROM game_winners
  GROUP BY match_id, round_id
  HAVING COUNT(*) * 2 > (
    SELECT COUNT(*)
      FROM games
      WHERE games.match_id = game_winners.match_id
      AND games.round_id = game_winners.round_id
  )

因此，一种选择是通过子查询添加一个检查，即玩家在比赛中赢得了超过一半的比赛。

SELECT games_won.match_id, games_won.round_id, games_won.player_id
  FROM (
    SELECT match_id, round_id, player_id, COUNT(*) AS win_count
      FROM game_winners
      GROUP BY match_id, round_id
    ) AS games_won
  INNER JOIN (
    SELECT match_id, round_id, COUNT(*) AS game_count
      FROM games
      GROUP BY match_id, round_id
    ) AS all_games
  ON games_won.match_id = game_count.match_id
  AND games_won.round_id = game_count.round_id
  WHERE games_won.win_count * 2 > all_games.game_count

或者，如果您愿意，可以使用联接来完成。

于 2011-08-07T20:21:24.327 回答