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我有一个名为 的表players，然后是其他名为tries, conversions, penalties,的表dropgoals。

我需要得到player_idfrom players，然后计算以下内容：

  • player_idin的尝试次数tries
  • player_idin的转化次数conversions
  • player_idin的处罚次数penalties
  • player_idin的丢球数dropgoals

这一切都需要一次性完成，因为每场比赛大约有十五名球员，该网站与橄榄球有关。

我尝试了以下方法，它有效：

SELECT players.player_id, players.number, CASE WHEN (COUNT(tries.player_id) = 0) THEN ' ' ELSE COUNT(tries.player_id) END AS nrTries FROM players LEFT JOIN tries ON players.player_id = tries.player_id WHERE players.team_id IS NULL GROUP BY players.player_id ORDER BY players.number

它从players表中选择所有玩家并计算他们的尝试，但是一旦我将其更改为以下内容，它就会给我一个错误：

SELECT players.player_id, players.number, player.name, CASE WHEN (COUNT(tries.player_id) = 0) THEN ' ' ELSE COUNT(tries.player_id) END AS nrTries FROM players, player LEFT JOIN tries ON players.player_id = tries.player_id WHERE players.player_id = player.player_id AND players.team_id IS NULL GROUP BY players.player_id ORDER BY players.number

它给出了以下错误：

Unknown column 'players.player_id' in 'on clause'

有人可以帮我解决这个问题吗，我已经挣扎了好几天了？

提前致谢

// 编辑：

大家好，我现在觉得很愚蠢，这段代码运行得很好，除了当一个球员在不止一种类型中得分时，比如尝试和转换，或者尝试和点球，它会错误计算每种类型的数量，并且它们是成功的都一样。

假设我的球员进了 1 个点球和 3 个丢球，它输出为 3 个点球和 3 个丢球，我不知道出了什么问题。

这是我的查询：

SELECT players.player_id, players.number, player.name, player.surname,
CASE WHEN (COUNT(tries.player_id) = 0) THEN '& nbsp;' ELSE COUNT(tries.player_id) END AS nrTries,
CASE WHEN (COUNT(conversions.player_id) = 0) THEN '& nbsp;' ELSE COUNT(conversions.player_id) END AS nrConversions,
CASE WHEN (COUNT(dropgoals.player_id) = 0) THEN '& nbsp;' ELSE COUNT(dropgoals.player_id) END AS nrDropgoals,
CASE WHEN (COUNT(penalties.player_id) = 0) THEN '& nbsp;' ELSE COUNT(penalties.player_id) END AS nrPenalties
FROM players
LEFT JOIN tries ON players.player_id = tries.player_id AND tries.game_id = '$game_id'
LEFT JOIN conversions ON players.player_id = conversions.player_id AND conversions.game_id = '$game_id'
LEFT JOIN dropgoals ON players.player_id = dropgoals.player_id AND dropgoals.game_id = '$game_id'
LEFT JOIN penalties ON players.player_id = penalties.player_id AND penalties.game_id = '$game_id'
LEFT JOIN player ON players.player_id = player.player_id
WHERE players.player_id = player.player_id AND players.team_id IS NULL AND players.game_id = '$game_id'
GROUP BY players.player_id ORDER BY players.number

请注意： $game_id 是一个 PHP 变量。
另外：我在 & 和 nbsp; 之间添加了一个空格。否则它不会输出到 SO。

有人可以指出我正确的方向吗？

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3 回答 3

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您对数据还不够具体，但是，假设每个玩家的每个关联表中有一条记录，并且如果当时不存在，您很乐意显示 NULL：

SELECT player_id, tries, conversions, penalties, dropgoals
FROM players p
LEFT JOIN tries t ON t.player_id = p.player_id
LEFT JOIN conversions c ON c.player_id = p.player_id
LEFT JOIN penalties e ON e.player_id = p.player_id
LEFT JOIN dropgoals d ON d.player_id = p.player_id

这可以重述为：

SELECT player_id
(SELECT tries FROM tries WHERE player_id = p.player_id) tries,
(SELECT conversions FROM conversions WHERE player_id = p.player_id) conversions,
(SELECT penalties FROM penalties WHERE player_id = p.player_id) penalties,
(SELECT dropgoals FROM dropgoals WHERE player_id = p.player_id) dropgoals
FROM players p

性能可能会也可能不会因您的数据库引擎而异。如果您需要对此求和，请将其更改为：

SELECT player_id
(SELECT SUM(tries) FROM tries WHERE player_id = p.player_id) tries,
(SELECT SUM(conversions) FROM conversions WHERE player_id = p.player_id) conversions,
(SELECT SUM(penalties) FROM penalties WHERE player_id = p.player_id) penalties,
(SELECT SUM(dropgoals) FROM dropgoals WHERE player_id = p.player_id) dropgoals
FROM players p

如果需要，以上任何一个都可以使用 IFNULL() 或类似函数返回 0 而不是 NULL。

于 2009-07-20T15:06:12.817 回答
1

我有点困惑为什么你说 Cletus 有正确的解决方案，并在更新你原来的问题时继续不使用它。也就是说，Cletus 的解决方案略有偏差，您需要 COUNT() 而不是 SUM()。尝试以下操作：

SELECT players.player_id,
(SELECT COUNT(*) FROM tries WHERE player_id = players.player_id) tries,
(SELECT COUNT(*) FROM penalties WHERE player_id = players.player_id) penalties
FROM players;

这将返回 0 而不是   我建议在您的应用程序代码中处理它。如果您真的想从 mysql 获取  ，您可以添加 CASE 混乱。

于 2009-07-21T16:17:33.183 回答
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为什么第二个查询引用“玩家”？我没有看到你在其他任何地方提到过这样的桌子......（“来自玩家，玩家”）只是确保这是故意的。

避免将显式连接（例如“LEFT JOIN”）与隐式连接（例如“FROM player, player”）混用可能会有所帮助，以保持结构一致。

于 2009-07-20T15:01:32.760 回答