将多个模块组合成一个可用的类-Python/SQLalchemy

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

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  

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