MySQL中的複雜足球聯盟動態排序？

Question

我有一個足球聯賽表「遊戲」如下：

date home_team_id away_team_id home_score  away_score 
-   1     2    6    21 
-   3     1    7    19 

我無法弄清楚如何動態地生成由勝利有序的團隊的ID列表（然後如果POSS點）？

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我有這個疑問時，我有一個$ TEAM_ID但事業的話，我只能做1隊的時間，並且不允許在查詢級別排序工作正常

((SELECT COUNT(*) FROM `games` WHERE ((`home_score` > `away_score`) AND `home_team_id` = '.$team_id.')) + 
(SELECT COUNT(*) FROM `games` WHERE ((`home_score` < `away_score`) AND `away_team_id` = '.$team_id.'))) AS `wins` 

我不知道我是否可以使用某種形式的GROUP，或者mySQL可以知道$ team_id本身？ 我也嘗試了一些與'團隊'表的多個JOIN，但他們也沒有工作。

感謝，

丹

Answer 1

也許這是你在找什麼呢？

SELECT all_wins.team_id, SUM(all_wins.wins) 
FROM (
    SELECT 
    home_team_id as team_id, 
    SUM(IF(home_score > away_score,1,0)) as wins, 
    SUM(home_score - away_score) as points 
    FROM games 
    GROUP BY home_team_id 
    UNION ALL 
    SELECT 
    away_team_id as team_id, 
    SUM(IF(away_score > home_score,1,0)) as wins, 
    SUM(away_score - home_score) as points 
    FROM games 
    GROUP BY away_team_id 
) all_wins 
GROUP BY all_wins.team_id 
ORDER BY SUM(all_wins.wins), SUM(all_wins.points) 

ETA：原來的答案並不完整，我認爲這應該會更好。

UNION在一起的內部兩個查詢獲得每個團隊的主客場勝利。外部查詢簡單地總結了主勝客的總勝率。

Answer 2

讓我們做一步一步來：

家選擇在家裏贏得比賽和比分：

SELECT COUNT(*) as wins, SUM(G.home_score) as score FROM games G WHERE 
     G.team_id = T.team_id #See 3. query and you'll understand 
     G.home_score > away_score 

讓我們把這個結果HOME_GAMES。

選擇贏得比賽和客場比賽的比分：

SELECT COUNT(*) as wins, SUM(G.away_score) as score FROM games G 
WHERE 
    G.team_id = T.team_id #See 3. query and you'll understand 
    G.away_score > G.home_score 

讓我們把這個結果AWAY_GAMES。

選擇總贏得比賽，總比分：

SELECT (A.wins + H.wins) AS total_wins, (A.score + H.score) AS total_score FROM 
    (AWAY_GAMES) AS A, (HOME_GAMES) AS H, teams T 
    ORDER BY total_wins, total_score 

==>代AWAY_GAMES和HOME_GAMES放在一起：

SELECT (A.wins + H.wins) AS total_wins, (A.score + H.score) AS total_score FROM 
    (SELECT COUNT(*) as wins, SUM(G.away_score) as score FROM games G 
    WHERE 
    G.team_id = T.team_id #See 3. and you'll understand 
    G.away_score > G.home_score) AS A, 

    (SELECT COUNT(*) as wins, SUM(G.home_score) as score FROM games G 
    WHERE 
     G.team_id = T.team_id #See 3. and you'll understand 
     G.home_score > away_score) AS H, 

    teams T 
    ORDER BY total_wins, total_score 

Answer 3

基於Eric的解決方案 - 這是我最後的查詢如果其他人有類似的問題 - 謝謝大家的幫助。

SELECT `teams`.`id`, `teams`.`name`, 
     SUM(`all_wins`.`gp`) AS `gp`, 
     SUM(`all_wins`.`w`) AS `w`, SUM(`all_wins`.`l`) AS `l`, SUM(`all_wins`.`t`) AS `t`, 
     SUM(`all_wins`.`ptf`) AS `ptf`, SUM(`all_wins`.`pta`) AS `pta` 
FROM (
    SELECT 
    `home_team_id` as `team_id`, 
    COUNT(`home_score`) AS `gp`, 
    SUM(IF(`home_score` > `away_score`,1,0)) as `w`, 
    SUM(IF(`home_score` < `away_score`,1,0)) as `l`, 
    SUM(IF(`home_score` = `away_score`,1,0)) as `t`, 
    SUM(IFNULL(`home_score`,0)) as `ptf`, 
    SUM(IFNULL(`away_score`,0)) as `pta` 
    FROM `games`  
    GROUP BY `home_team_id` 
    UNION ALL 
    SELECT 
    `away_team_id` as `team_id`, 
    COUNT(`home_score`) AS `gp`, 
    SUM(IF(`away_score` > `home_score`,1,0)) as `w`, 
    SUM(IF(`away_score` < `home_score`,1,0)) as `l`, 
    SUM(IF(`away_score` = `home_score`,1,0)) as `t`, 
    SUM(IFNULL(`away_score`,0)) as `ptf`, 
    SUM(IFNULL(`home_score`,0)) as `pta` 
    FROM `games` 

    GROUP BY `away_team_id` 
) `all_wins` 
LEFT JOIN `teams` ON `all_wins`.`team_id` = `teams`.`id` 
GROUP BY `all_wins`.`team_id` 
ORDER BY SUM(`all_wins`.`w`) DESC, SUM(`all_wins`.`ptf`) DESC