mmc*_*han 13 python sqlalchemy
我在同一个模块中有两个模型models.它们是1-1关系,并且已根据SQLAlchemy文档进行配置.
Vehicle.py
from models.AssetSetting import AssetSetting
class Vehicle(Base):
__tablename__ = 'vehicles'
vehicle_id = Column(Integer, primary_key=True)
...
settings = relationship('AssetSetting', backref=backref('asset_settings'))
Run Code Online (Sandbox Code Playgroud)
AssetSetting.py
from models.Vehicle import Vehicle
class AssetSetting(Base):
__tablename__ = 'asset_settings'
asset_alert_setting_id = Column(Integer, primary_key=True, autoincrement=True)
...
vehicle = relationship('vehicles', foreign_keys=Column(ForeignKey('vehicles.vehicle_id')))
Run Code Online (Sandbox Code Playgroud)
如果我使用字符串关系构建(即ForeignKey('vehicles.vehicle_id'))我得到错误:
sqlalchemy.exc.InvalidRequestError:
When initializing mapper Mapper|AssetSetting|asset_settings, expression 'vehicles' failed to locate a name ("name 'vehicles' is not defined").
If this is a class name, consider adding this relationship() to the <class 'models.AssetSetting.AssetSetting'> class after both dependent classes have been defined.
Run Code Online (Sandbox Code Playgroud)
如果我使用类映射,我会得到经典的循环导入错误:
Traceback (most recent call last):
File "tracking_data_runner.py", line 7, in <module>
from models.Tracker import Tracker
File "/.../models/Tracker.py", line 5, in <module>
from models.Vehicle import Vehicle
File "/.../models/Vehicle.py", line 13, in <module>
from models.Tracker import Tracker
ImportError: cannot import name 'Tracker'
Run Code Online (Sandbox Code Playgroud)
我相信我可以通过将文件放在同一个包中来解决这个问题,但我更愿意将它们分开.思考?
ale*_*cxe 11
为了避免循环导入错误,您应该使用字符串关系构建,但是两个模型都必须使用Base相同的declarative_base实例.实例化Base一次初始化时都用它Vehicle和AssetSetting.
或者,您可以显式映射表名和类,以帮助映射器关联您的模型:
Base = declarative_base(class_registry={"vehicles": Vehicle, "asset_settings": AssetSetting})
Run Code Online (Sandbox Code Playgroud)
mmc*_*han 11
我发现我的问题有两个问题:
Vehicles在我的关系中不恰当地引用.应该relationship('Vehicle'不是relationship('vehicles'foreign_keys=Column(ForeignKey('vehicles.vehicle_id')))中那样在关系中声明FK是不合适的.我必须声明FK,然后将其传递给关系.我的配置现在看起来像这样:
Vehicle.py
class Vehicle(Base, IDiagnostable, IUsage, ITrackable):
__tablename__ = 'vehicles'
vehicle_id = Column(Integer, primary_key=True)_id = Column(Integer)
settings = relationship('AssetSetting', backref=backref('asset_settings'))
Run Code Online (Sandbox Code Playgroud)
AssetSetting.py
class AssetSetting(Base):
__tablename__ = 'asset_settings'
asset_alert_setting_id = Column(Integer, primary_key=True, autoincrement=True)
vehicle_id = Column(ForeignKey('vehicles.vehicle_id'))
vehicle = relationship('Vehicle', foreign_keys=vehicle_id)
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
5518 次 |
| 最近记录: |