Question

我正在尝试定义两个关系是间接关系的表之间的关系（即通过另外两个表）。

我正在寻找的结果可以使用此查询获取：

(db.session.query(Telnum)
           .filter(Account.customer==customer)
           .filter(Account.account_id == Subscription.account_id)
           .filter(Telnum.sub_id == Subscription.id)
           .order_by(Telnum.telnum)
           .all()
)

其中customer是Customer对象。

我正在努力弄清楚如何定义这种关系，类似于Customer.invoices关系。我的想法是这样的：

telnums = db.relationship('Telnum',
                       primaryjoin="and_(Account.user_id==Customer.id, "
                       "Account.account_id == Subscription.account_id, " 
                       "Telnum.sub_id == Subscription.id)", 
                       backref='customer')

这篇文章很明显，这不起作用。它产生的错误信息是： sqlalchemy.exc.ArgumentError: Could not locate any simple equality expressions involving locally mapped foreign key columns for primary join condition 'accounts.user_id = customers.id AND accounts.account_id = subscriptions.account_id AND pstn_numbers.sub_id = subscriptions.id' on relationship Customer.telnums. Ensure that referencing columns are associated with a ForeignKey or ForeignKeyConstraint, or are annotated in the join condition with the foreign() annotation. To allow comparison operators other than '==', the relationship can be marked as viewonly=True.

有人能向我暗示正确的方向吗？

我有以下表结构（简化，除了每个表的一个之外，所有不相关的列都被删除）：

class Customer(db.Model):
    __tablename__ = 'customers'
    id = db.Column(db.Integer, primary_key=True)
    identification_num = db.Column(db.String(10), unique=True)
    name = db.Column(db.Text)
    invoices = db.relationship('Invoice', backref='customer')
    accounts = db.relationship('Account', backref='customer')

def __init__(self):
    pass

def __repr__(self):
    return '<Customer %r>' % (self.name)

class Invoice(db.Model):
    __tablename__ = 'invoices'
    id = db.Column(db.Integer, primary_key=True)
    customer_id = db.Column(db.Integer, db.ForeignKey('customers.id'))
    active = db.Column(db.Boolean)
    accounts = db.relationship('Account', backref='invoice')

    def __repr__(self):
        return '<Invoice %r>' % (self.id)

class Account(db.Model):
    __tablename__ = 'accounts'
    id = db.Column(db.Integer, primary_key=True)
    account_id = db.Column(db.Integer, unique=True)
    invoice_id = db.Column(db.Integer, db.ForeignKey('invoices.id'))
    user_id = db.Column(db.Integer, db.ForeignKey('customers.id'))
    active = db.Column(db.Boolean)
    subscriptions = db.relationship('Subscription', backref='account')

    def __repr__(self):
        return '<Account %r>' % (self.account_id)

class Subscription(db.Model):
    __tablename__ = 'subscriptions'
    id = db.Column(db.Integer, primary_key=True)
    account_id = db.Column(db.Integer, db.ForeignKey('accounts.account_id'))
    sub_active = db.Column(db.DateTime)
    telnums = db.relationship('Telnum', backref='subscription')

    def __repr__(self):
        return '<Subscription %r>' % (self.id)

class Telnum(db.Model):
    __tablename__ = 'pstn_numbers'
    id = db.Column(db.Integer, primary_key=True)
    sub_id = db.Column(db.Integer, db.ForeignKey('subscriptions.id'))
    telnum = db.Column(db.String(64))
    holder = db.Column(db.String(10))

    def __repr__(self):
        return '<Telnum %r>' % (self.telnum)

Answer 1

一般情况下，我不会将间接关系定义为relationship，因为当您制作时，这些间接关系会变得不同步修改。您可以通过为relationship指定viewonly=False参数来解决其中一些限制。

更简单，风险更小，更直接的解决方案是使用查询（或启用查询的属性），以防您想从数据库重新加载数据，并使用python list comprehensions来获取子关系树的子孩子：

class Customer(Base):
    # ...

    @property
    def telnums_qry(self):
        sess = Session.object_session(self)
        return (sess.query(Telnum)
                .join(Subscription)
                .join(Account)
                .filter(Account.user_id == self.id)
                ).all()


    @property
    def telnums_mem(self):
        return [tel
                for acc in self.accounts
                for sub in acc.subscriptions
                for tel in sub.telnums
                ]


class Telnum(Base):
    # ...

    @property
    def customer(self):
        return (self.subscription
                and self.subscription.account
                and self.subscription.account.customer
                )

如何在SQLAlchemy中使用多级/多连接指定表关系？

1 个答案: