Динамично подреждане на сложна футболна лига в MySQL?

Нека го направим стъпка по стъпка:

Изберете спечелените игри у дома и резултата у дома:

   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

Нека наречем този резултат HOME_GAMES.

Изберете спечелените игри и резултата от гостуванията:

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

Нека наречем този резултат AWAY_GAMES.

Изберете общия брой спечелени игри и общия резултат:

   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

==> Съберете всичко, като замените AWAY_GAMES и HOME_GAMES:

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