将多个模块组合成一个可用的类-Python/SQLalchemy
我有多个表(和计数)模型模块,我希望类“table”继承它们。代码如下(简化)。当我向我的项目中添加更多的表模型模块时,是否有任何方法可以使其动态化?现在看来,我必须在“表”类导入和构造函数中显式列出每个模块将多个模块组合成一个可用的类-Python/SQLalchemy,python,inheritance,dynamic,sqlalchemy,simplify,Python,Inheritance,Dynamic,Sqlalchemy,Simplify,我有多个表(和计数)模型模块,我希望类“table”继承它们。代码如下(简化)。当我向我的项目中添加更多的表模型模块时,是否有任何方法可以使其动态化?现在看来,我必须在“表”类导入和构造函数中显式列出每个模块 from app.database.model_Ingredient_Inventory import Ingredient_Inventory from app.database.model_Ingredient_Remaining import Ingredient_Remaining
from app.database.model_Ingredient_Inventory import Ingredient_Inventory
from app.database.model_Ingredient_Remaining import Ingredient_Remaining
class Table:
def __init__(self):
self.Ingredient_Inventory = Ingredient_Inventory
self.Ingredient_Remaining = Ingredient_Remaining
其他两个表格模型模块(简化)。单元1:
from app import db
class Ingredient_Inventory(db.Model):
__tablename__ = 'Ingredient_Inventory'
id = db.Column(db.BigInteger, primary_key=True)
ingredientID = db.Column(db.BigInteger, unique = False)
def __init__(self, ingredientID, ingredientTypeID)
self.ingredientID = ingredientID
self.ingredientTypeID = ingredientTypeID
单元2:
from app import db
class Ingredient_Remaining(db.Model):
__tablename__ = 'Ingredient_Remaining'
id = db.Column(db.BigInteger, primary_key=True)
currentValue = db.Column(db.BigInteger, unique = False)
def __init__(self, currentValue):
self.currentValue = currentValue