如何用SQLAlchemy求和* count *子查询？

Question

我有以下型号在我的DB（瓶-SQLAlchemy中，声明的方式，简化）：

class Player(db.Model): 
    id = db.Column(db.Integer, primary_key = True) 
    ... 

class Game(db.Model): 
    id = db.Column(db.Integer, primary_key = True) 
    creator_id = db.Column(db.Integer, db.ForeignKey('player.id')) 
    creator = db.relationship(Player, foreign_keys='Game.creator_id') 
    opponent_id = db.Column(db.Integer, db.ForeignKey('player.id')) 
    opponent = db.relationship(Player, foreign_keys='Game.opponent_id') 
    winner = db.Column(db.Enum('creator', 'opponent')) 

每场比赛可以是韩元，丢失或平局收场。 我需要通过“赢率”为了让玩家对它们进行排序 - 即：

• 如果玩家创建了一个游戏，那场比赛的获胜者creator，它被认为是双赢;
• 如果玩家被邀请参加比赛作为对手并且游戏的获胜者是opponent，那么它也被视为赢;
• 此玩家参与的其他游戏被视为丢失游戏。

所以我的算法如下：

@hybrid_property 
def winrate(self): 
    games = Game.query.filter(or_(
     Game.creator_id == self.id, 
     Game.opponent_id == self.id, 
    )) 
    count = 0 
    wins = 0 
    for game in games: 
     count += 1 
     if game.creator_id == self.id and game.winner == 'creator': 
      wins += 1 
     elif game.opponent_id == self.id and game.winner == 'opponent': 
      wins += 1 
    if count == 0: 
     return 0 
    return wins/count 

，这个方法工作时，我想确定赢率特定的球员;但是当我想按胜率排序时，它就失败了。 我试图重写它的SQL和得到的东西是这样的：

SELECT * FROM player 
ORDER BY ((SELECT count(g1.id) FROM game g1 
    WHERE g1.creator_id = player.id AND g1.winner = 'creator' 
) + (SELECT count(g2.id) FROM game g2 
    WHERE g2.opponent_id = player.id AND g2.winner = 'opponent' 
))/(SELECT count(g3.id) FROM game g3 
    WHERE player.id IN (g3.creator_id, g3.opponent_id) 
) 

这不处理球员没有比赛，但一般应工作。没有游戏的玩家可以用MySQL CASE声明处理。

但问题是我无法确定如何使用SQLAlchemy编码此SQL。 这里是（简化）代码，我尝试使用：

@winrate.expression 
def winrate(cls): 
    cnt = Game.query.filter(
     cls.id.in_(Game.creator_id, Game.opponent_id) 
    ).with_entities(func.count(Game.id)) 
    won = Game.query.filter(
     or_(
      and_(
       Game.creator_id == cls.id, 
       Game.winner == 'creator', 
      ), 
      and_(
       Game.opponent_id == cls.id, 
       Game.winner == 'opponent', 
      ), 
     ) 
    ) 
    return case([ 
     (count == 0, 0), 
    ], else_ = (
     won/count 
    )) 

当谈到won/count线告诉我，Query不能Query分割该代码失败。我尝试使用子查询，但没有任何成功。

我应该如何实现它？或者，也许我应该使用某种连接/任何？ （DB计划不能更改。）

Answer 1

尝试用核表达，而不是ORM查询工作：

class Player(..): 
    # ... 
    @winrate.expression 
    def _winrate(cls): 
     cnt = (
      select([db.func.count(Game.id)]) 
      .where(
       db.or_(
        Game.creator_id == cls.id, 
        Game.opponent_id == cls.id, 
       )) 
      .label("cnt") 
     ) 
     won = (
      select([db.func.count(Game.id)]) 
      .where(
       db.or_(
        db.and_(Game.creator_id == cls.id, 
          Game.winner == 'creator'), 
        db.and_(Game.opponent_id == cls.id, 
          Game.winner == 'opponent'), 
       )) 
      .label("cnt") 
     ) 

     return db.case(
      [(cnt == 0, 0)], 
      else_ = db.cast(won, db.Numeric)/cnt 
     ) 
# ... 
q = session.query(Player).order_by(Player.winrate.desc())