sqlalchemy hybrid_property和表达式

Question

我不能理解这些表达。如何让以下代码正常工作？

class OperationType(Enum):
    MINUS = 1
    MINUS_CORR = 2
    PLUS = 3
    PLUS_CORR = 4

按类型分组操作

BALANCE_PLUS_OPERATIONS = [
    OperationType.PLUS.value,
    OperationType.PLUS_CORR.value
]

BALANCE_MINUS_OPERATIONS = [
    OperationType.MINUS.value,
    OperationType.MINUS_CORR.value
]

运营模式

class Operation(Model):

    __tablename__ = 'operation'

    id = db.Column(db.BigInteger, primary_key=True)
    created_at = Column(db.DateTime, nullable=False, default=dt.datetime.utcnow)
    operation_type = db.Column(db.SmallInteger, nullable=False)
    amount = Column(db.Integer, nullable=False)
    user_id = db.Column(db.ForeignKey('users.id'), nullable=False)
    user = relationship('User', backref='operation', uselist=False)

用户模型

class User(UserMixin, Model):

    __tablename__ = 'users'

    id = Column(db.Integer, primary_key=True)
    operations = relationship("Operation", backref="users")

    @hybrid_property
    def balance(self):
        plus = sum(op.amount for op in self.operations if op.operation_type in BALANCE_PLUS_OPERATIONS)
        minus = sum(op.amount for op in self.operations if op.operation_type in BALANCE_MINUS_OPERATIONS)
        return plus - minus

    @balance.expression
    def balance(cls):
        p = select([func.sum(Operation.amount).label('BALANCE_PLUS_OPERATIONS')]) \
                .where(Operation.operation_type.in_(BALANCE_PLUS_OPERATIONS)) \
                .where(User.id == cls.id) \
                .as_scalar()
        m = select([func.sum(Operation.amount).label('BALANCE_MINUS_OPERATIONS')]) \
                .where(Operation.operation_type.in_(BALANCE_MINUS_OPERATIONS)) \
                    .where(User.id == cls.id) \
                    .as_scalar()
        return select([p - m]).label('BALANCE')

表达式是错误的，并且将产生错误的结果：

users = User.query.filter_by(balance=51).all()
for u in users:
    print(u, u.balance)

打印：

<User(foo@bar.com)> 51
<User(bar@foor.com)> 0

但我只期望有一条记录：

<User(foo@bar.com)> 51

谢谢

Answer 1

我将从上下文中假设这些方法属于User类。在这样的光芒下

.where(User.id == cls.id) \

是有效的

.where(User.id == User.id) \

或者仅仅是where(True)，所以每个用户都会加入到每个操作中，而这可能意味着像这样

.where(Operation.user_id == cls.id) \

虽然由于缺乏实例而不能说。如果确实发生了不正确的联接，它解释了查询返回另一个用户的原因:它是通过属于正确用户的操作联接的。

您可能还需要添加

.correlate(cls) \

在as_scalar()之前。我认为最外层的select也是多余的。你应该能够

return (p - m).label('BALANCE')