MySQL - 需要幫助計數id與其他表中的某些值相對應

Question

我在表上命名爲players，然後其他表名爲tries，conversions,penalties, dropgoals。在conversions

數量的處罰爲player_id在tries

數轉換爲player_id的

• 嘗試次數爲player_id的：

  我需要從players的player_id，再算上以下penalties

• 下降目標數player_id in dropgoals

而這一切都需要一次性完成，因爲每場比賽都有大約15名球員，該網站涉及橄欖球。

我曾嘗試以下，其工作原理：

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

它選擇從players表中的所有玩家，並計算他們的嘗試，但只要我把它更改爲以下它給了我一個錯誤：

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

它提供了以下錯誤：

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

有人可以幫我這個，我一直在掙扎幾天嗎？

在此先感謝

//編輯：

嗨，好，我現在感覺非常愚蠢，此代碼的工作出色，除了當一個球員在多於一個類型已經打進了，說嘗試和轉換，或嘗試和處罰，它錯誤地計數每種類型的數量，並且它們都是一樣的。假設我的球員獲得了1罰則和3個dropgoals，輸出爲3罰分和3個dropgoals，我無法弄清楚什麼是錯誤的。

這裏是我的查詢：

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 

請注意： $ game_id是一個PHP變量。
另外：我在&和nbsp;否則它不會被輸出到SO。

有人能請我指出正確的方向嗎？

Answer 1

我有點困惑，爲什麼你說Cletus有正確的解決方案，並且在更新原始問題時不使用它。也就是說，Cletus的解決方案稍微關閉，你需要COUNT（）而不是SUM（）。請嘗試以下操作：

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; 

這將返回0的，而不是& NBSP的我建議你處理，在應用程序代碼，但。如果你真的想要從mysql獲取，你可以在CASE亂七八糟的地方添加。

Answer 2

爲什麼第二個查詢引用「player」？我沒有看到你在其他地方提過這樣的桌子......（「從玩家，玩家」）只要確保這是故意的。

爲了保持結構的一致性，避免混合使用隱式連接（例如「FROM players，player」）的顯式連接（例如「LEFT JOIN」）可能會有所幫助。

Answer 3

您還沒有有關數據不夠具體，但是，假設有在每個玩家每個相關表中的一個記錄和你快樂顯示NULL，如果它不存在，那麼：

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 

這可以重申爲：

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 

性能可能會也可能不會因您的數據庫引擎而異。如果您需要總結這則改爲：

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 

上述任何可以使用IFNULL（）或類似的功能，返回0，而不是NULL，如果需要的話。