Question

假设我有一个类似这样的SqlAlchemy模型：

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, String, Integer, ForeignKey
from sqlalchemy.orm import sessionmaker, relationship
Base = declarative_base()
Session = sessionmaker()

class EmployeeType(Base):
    __tablename__ = 'employee_type'
    id = Column(Integer(), primary_key=True)
    name = Column(String(20))

class Employee(Base):
    __tablename__ = 'employee'
    id = Column(Integer(), primary_key=True)
    type_id = Column(Integer(), ForeignKey(EmployeeType.id))
    type = relationship(EmployeeType, uselist=False)

session = Session()
session.add(EmployeeType(name='drone'))
session.add(EmployeeType(name='PHB'))

我想要有某种关系＆＃34;从Employee直接到EmployeeType.name为方便，所以如果我有一个类型名称，我可以跳过查找id或EmployeeType对象的步骤：

emp = Employee()
emp.type_name = "drone"
session.add(emp)
session.commit()
assert (emp.type.id == 1)

这样的事情可能吗？

编辑：我发现association_proxy可以让我在那里：

class Employee(Base):
    ...
    type_name = association_proxy("type", "name")

唯一的问题是，如果我分配给它：

emp = session.query(Employee).filter_by(EmployeeType.name=='PHB').first()
emp.type_name = 'drone'

它修改employee_type.name列，而不是employee.type_id列。

Answer 1

我会通过创建一个为我这样做的方法来做到这一点。

class EmployeeType(Base):
    __tablename__ = 'employee_type'
    id = Column(Integer(), primary_key=True)
    name = Column(String(20))

class Employee(Base):
    __tablename__ = 'employee'
    id = Column(Integer(), primary_key=True)
    type_id = Column(Integer(), ForeignKey(EmployeeType.id))
    type = relationship(EmployeeType, uselist=False)

    def __init__(self, type):
        self.type = type

    def add(self, type_name=None):
        if type_name is not None:
            emp_type = DBSession.query(EmployeeType).filter(EmployeeType.name == type_name).first()
            if emp_type:
                type = emp_type
            else:
                type = EmployeeType(name=type_name)
        else:
            type = None
        DBSession.add(Employee(type=type))

然后你做：

Employee.add(type_name='boss')

Answer 2

我同意Jonathan的一般方法，但我想在会话中添加一个员工对象，并且设置员工类型应该是独立的操作。这是一个将type_name作为属性的实现，需要在设置之前添加到会话：

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, String, Integer, ForeignKey
from sqlalchemy.orm import sessionmaker, relationship
Base = declarative_base()
Session = sessionmaker()

class EmployeeType(Base):
    __tablename__ = 'employee_type'
    id = Column(Integer(), primary_key=True)
    name = Column(String(20))

class Employee(Base):
    __tablename__ = 'employee'
    id = Column(Integer(), primary_key=True)
    type_id = Column(Integer(), ForeignKey(EmployeeType.id))
    type = relationship(EmployeeType)

    @property
    def type_name(self):
        if self.type is not None:
            return self.type.name
        return None

    @type_name.setter
    def type_name(self, value):
        if value is None:
            self.type = None
        else:
            session = Session.object_session(self)
            if session is None:
                raise Exception("Can't set Employee type by name until added to session")
            self.type = session.query(EmployeeType).filter_by(name=value).one()

SqlAlchemy与特定列的关系

2 个答案: