sqlalchemy hybrid_property和表达式

Question

I can not understand the expressions. How to get the following code to work?

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

group operation by type

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

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

Operation model

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)

User model

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')

The expression is wrong and will produce the wrong result:

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

print:

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

but i expected only one record:

<User(foo@bar.com)> 51

Thanks

Answer 1

I'll assume from context that the methods belong to the User class. In that light

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

is effectively

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

or just where(True) , and so every user is joined with every operation, when it probably is meant to be something like

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

though impossible to say due to lacking example. If the incorrect join does happen, it explains why the other user is returned by the query: it was joined with operations belonging to the correct user.

You also might have to add

.correlate(cls) \

before as_scalar() . The outermost select is also redundant, I think. You should be able to just

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