Question

这是我的问题的精简版：

考虑我有两个表：'procedure'和'role'。

role有字段：（role_uid，role_name）

程序包含字段：（procedure_uid，procedure_name，inform_role_uid，consult_role_uid）

所以'角色'与'程序'有两个一对多的关系。

一些代码：

class Role(Base):
    __tablename__ = "role"
    __table_args__ = ({'autoload':True, 'useexisting': True})

class Procedure(Base):
    __tablename__ = "procedure"
    __table_args__ = (sqlalchemy.ForeignKeyConstraint(['consult_role_uid','inform_role_uid'],['role.role_uid', 'role.role_uid']),
        {'autoload':True, 'useexisting': True})

Procedure.consult_role = sqlalchemy.orm.relationship(Role,
 primaryjoin="Procedure.consult_role_uid==Role.role_uid", foreign_keys=Role.role_uid)
Procedure.inform_role = sqlalchemy.orm.relationship(Role,
 primaryjoin="Procedure.inform_role_uid==Role.role_uid", foreign_keys=Role.role_uid)

consult_role = sqlalchemy.orm.aliased(Role, name="consult_role")
inform_role = sqlalchemy.orm.aliased(Role, name="inform_role")

query = session.query(
    Procedure.procedure_name, 
    consult_role.role_name.label("consult_role_name"),
    inform_role.role_name.label("inform_role_name")).join(consult_role, inform_role)

这会生成以下SQL：

SELECT 
  `procedure`.procedure_name AS procedure_procedure_name, 
  consult_role.role_name AS consult_role_name, 
  inform_role.role_name AS inform_role_name 
FROM 
  `procedure` 
  INNER JOIN role AS consult_role 
    ON consult_role.role_uid = `procedure`.consult_role_uid
      AND consult_role.role_uid = `procedure`.inform_role_uid 
  INNER JOIN role AS inform_role 
    ON inform_role.role_uid = `procedure`.consult_role_uid 
      AND inform_role.role_uid = `procedure`.inform_role_uid

正如您所看到的，我无意在每个字段上加入每个内部联接。为什么它似乎忽略了我的'primaryjoin'论点？

Answer 1

此：

sqlalchemy.ForeignKeyConstraint(['consult_role_uid','inform_role_uid'],['role.role_uid', 'role.role_uid'])

粗略地说，两个表之间的关系是通过两个属性来表达的，就像referant有一个复合主键一样。如果要有两个外键引用，则需要指定ForeignKey两次。

class Procedure(Base):
    __tablename__ = "procedure"
    __table_args__ = (
        sqlalchemy.ForeignKeyConstraint(['consult_role_uid'],['role.role_uid']),
        sqlalchemy.ForeignKeyConstraint(['inform_role_uid'],['role.role_uid']),
        {'autoload':True, 'useexisting': True})

Answer 2

因此，为了完整性，以下是上述问题的固定代码。我添加了两个ForeignKeyContstaints，我还必须指定在连接中使用哪种关系。

class Role(Base):
    __tablename__ = "role"
    __table_args__ = ({'autoload':True, 'useexisting': True})


class Procedure(Base):
    __tablename__ = "procedure"
    __table_args__ = (
        sqlalchemy.ForeignKeyConstraint(['consult_role_uid'], ['role.role_uid']),
        sqlalchemy.ForeignKeyConstraint(['inform_role_uid'], ['role.role_uid']),
        {'autoload':True, 'useexisting': True})

Procedure.consult_role = sqlalchemy.orm.relationship(Role,
 primaryjoin="Procedure.consult_role_uid==Role.role_uid", foreign_keys=Role.role_uid)
Procedure.inform_role = sqlalchemy.orm.relationship(Role,
 primaryjoin="Procedure.inform_role_uid==Role.role_uid", foreign_keys=Role.role_uid)

consult_role = sqlalchemy.orm.aliased(Role, name="consult_role")
inform_role = sqlalchemy.orm.aliased(Role, name="inform_role")

query = session.query(
    Procedure.procedure_name, 
    consult_role.role_name.label("consult_role_name"),
    inform_role.role_name.label("inform_role_name")).join((consult_role, Procedure.consult_role), (inform_role, Procedure.inform_role))

这产生了以下正确的SQL：

SELECT 
  `procedure`.procedure_name AS procedure_procedure_name, 
  consult_role.role_name AS consult_role_name, 
  inform_role.role_name AS inform_role_name 
FROM 
  `procedure` 
  INNER JOIN role AS consult_role ON `procedure`.consult_role_uid = consult_role.role_uid     
  INNER JOIN role AS inform_role ON `procedure`.inform_role_uid = inform_role.role_uid

sqlalchemy：使用声明和反射多次连接到同一个表

2 个答案: