Flask-SQLAlchemy连接3个模型和一个Table构造

Question

我有3个型号:

class Customer(Model):
    __tablename__ = 'customer'

    id = Column(Integer, primary_key=True)
    statemented_branch_id = Column(Integer, ForeignKey('branch'))
    ...

class Branch(Model):
    __tablename__ = 'branch'

    id = Column(Integer, primary_key=True)
    ...

class SalesManager(Model):
    __tablename__ = 'sales_manager'

    id = Column(Integer, primary_key=True)
    branches = relationship('Branch', secondary=sales_manager_branches)

和一个表构造:

sales_manager_branches = db.Table(
    'sales_manager_branches',
    Column('branch_id', Integer, ForeignKey('branch.id')),
    Column('sales_manager_id', Integer, ForeignKey('sales_manager.id'))
)

我希望能够得到所有Customers的SalesManager,这意味着所有有客户statemented_branch_id在任何的BranchES的SalesManager.branches关系.

我的查询看起来有点像这样:

branch_alias = aliased(Branch)
custs = Customer.query.join(branch_alias, SalesManager.branches).\
        filter(Customer.statemented_branch_id == branch_alias.id)

这显然是不对的.

我怎么能得到Customers一个SalesManager？

更新

当我尝试:

Customer.query.\
         join(Branch).\
         join(SalesManager.branches).\
         filter(SalesManager.id == 1).all()

我得到一个OperationalError:

*** OperationalError: (OperationalError) ambiguous column name: branch.id u'SELECT
customer.id AS customer_id, customer.statemented_branch_id AS 
customer_statemented_branch_id \nFROM customer JOIN branch ON branch.id 
customer.statemented_branch_id, "SalesManager" JOIN sales_manager_branches AS 
sales_manager_branches_1 ON "SalesManager".id = sales_manager_branches_1.sdm_id JOIN 
branch ON branch.id = sales_manager_branches_1.branch_id \nWHERE "SalesManager".id = ?'
(1,)

Answer 1

我需要在backref我的SalesManager模型中添加一个允许SQLAlchemy弄清楚如何从SalesManager分支到达的模型.

class SalesManager(Model):
    __tablename__ = 'sales_manager'

    id = Column(Integer, primary_key=True)
    branches = relationship(
        'Branch', secondary=sales_manager_branches, backref="salesmanagers")

并构造这样的查询:

Customer.query.\
         join(Branch).\
         join(Branch.salesmanagers).\
         filter(SalesManager.id == 1).all()

Answer 2

尝试：

SalesManager.query \
            .join(Branch) \
            .join(Customer) \
            .filter(SalesManager.id == 123)

您可能需要on通过第二个参数提供显式参数，join或者您可能需要显式添加映射表 - 但在任何一种情况下，您想要执行的操作如下：

SELECT SM.*
FROM sales_manager SM
JOIN sales_manager_branches SMB
    ON SM.id = SMB.sales_manager_id
JOIN branch B
    ON SMB.branch_id = B.id
JOIN customer C
    ON B.id = C.statemented_branch_id
WHERE -- Conditions go here