6

我有 3 个模型：

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)

和一个表结构：

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Customers为 a获得全部SalesManager，这意味着所有statemented_branch_id在任何BranchesSalesManager.branches关系中都有 a 的客户。

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。

我怎样才能得到Customers一切SalesManager？

更新

当我尝试：

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

我得到一个操作错误：

*** 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,)
4

2 回答 2

8

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分支到分支的方法。

class SalesManager(Model):
    __tablename__ = 'sales_manager'

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

并像这样构造查询：

Customer.query.\
         join(Branch).\
         join(Branch.salesmanagers).\
         filter(SalesManager.id == 1).all()
于 2013-07-23T07:12:56.483 回答
1

尝试：

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

您可能需要on通过 的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 - 但无论哪种情况，您都尝试执行以下操作：

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
于 2013-07-23T05:57:42.737 回答