如何使用 SQLAlchemy 对 count 个子查询求和？

Question

我的数据库中有以下模型（Flask-SQLALchemy，声明式方法，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'))

每场比赛可能赢、输或以平局结束。我需要让玩家按 "win rate" 对它们进行排序 - 即：

如果玩家创建了一个游戏并且该游戏的获胜者是creator，则视为获胜；
如果玩家作为对手被邀请参加游戏，并且游戏的获胜者是opponent，也算赢；
该玩家参加的其他游戏被视为输掉的游戏。

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

当我想确定特定玩家的胜率时，这种方法很有效；但是当我想按胜率对玩家进行排序时它失败了。我试图在 SQL 中重写它并得到类似这样的东西：

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)
)

这不适用于没有游戏的玩家，但一般情况下应该可以。没有游戏的玩家可能可以用 MySQL CASE 语句来处理。

但问题是我不知道如何使用 SQLAlchemy 对 SQL 进行编码。这是我尝试使用的（简化的）代码：

@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
    ))

此代码在 won / count 行时失败，告诉我 Query 不能被 Query 除。我尝试使用子查询但没有成功。

我该如何实施？或者我应该使用某种 joins/whatever？（无法更改数据库方案。）

Answer 1

尝试使用核心表达式而不是 orm 查询：

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())

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