问题描述
假设我有一个模型，员工：
员工
- 名称：
CharField - 电子邮件：
EmailField - 类型：
ForeignKey
- Name:
CharField - Email:
EmailField - Type:
ForeignKey
问题出在 type 字段。我希望外键能够几种不同类型的模型。例如，一个用于 Software Developer ，另一个用于 Janitor ，依此类推。
The problem is with the type field. I want the foreign key to be able to be several different types of models. For example, one for Software Developer, another for Janitor, and so on.
我可以通过为所有类型设置一个外键字段并将其中许多设置为null来做到这一点，但这似乎很糟糕。
如何实现？谢谢！！！！
How can I achieve this? Thanks!!
推荐答案
有很多方法可以将这种多态水平建模到这样的场景中。
There are many ways you can model this level of polymorphism into a scenario like this.
选项1-混凝土基础模型
OPTION 1 - Concrete Base Model
包括 employee_type 字段，指示正在使用的具体模型。
Including a employee_type field which indicates the concrete model being used.
class Employee(models.Model):
name = models.CharField(max_length=50)
email = models.CharField(max_length=80)
employee_type = models.CharField(max_length=20)
class Janitor(Employee):
[...]
class SoftwareEngineer(Employee):
[...]
所有共同属性都可以存储在 Employee 模型中。
All mutual attributes can be stored in the Employee model.
一次具体类（软件工程师，Janitor等）被实例化，它将继承其父类的所有属性。
Once a concrete class (Software Engineer, Janitor etc.) is instantiated it will inherit all attributes from it's parent class.
设置并使用 employee_type 您可以区分下注
Setting and using the employee_type you can differentiate between which concrete class was created.
有关此内容的更多信息，请参见
More information on this can be found here
[更新]
使用Django 信号可以派生具体的类名称，并将其与关联的实例一起存储。
Using a django Signal the concrete class name can be derived and stored with the associated instance.
signals.py
from django.db.models.signals import pre_save
def store_classname(sender, instance, **kwargs):
instance.employee_type = instance.__class__.__name__
for subclass in Employee.__subclasses__():
pre_save.connect(store_classname, sender=subclass)
这可确保每次都存储正确的标识符。
This ensures the correct identifier is stored every time.
现在在您的视图中要选择具体类的类型使用过，您可以在以下情况下继续使用它：
Now in your view where you want to select the type of concrete class to be used, you can either continue using it in a condition like so:
views.py
#EXAMPLE USAGE
employee = Employee.objects.get(id=1)
if employee.employee_type == 'Janitor':
employee = Janitor.objects.get(id=employee.id)
或通过使用模型名称和对全局变量的查找来动态派生它。
Or derive it dynamically by using the model name and a lookup to global variables.
#EXAMPLE USAGE
from <you_app>.models import *
def get_class(classname):
cls = globals()[classname]
return cls
employee = Employee.objects.get(id=1)
concrete_employee = get_class(employee.employee_type)
[...]
注意：更改父模型或子模型的名称时要小心，因为这会影响使用旧模型名称的历史记录。要解决此问题，请使用django的 update（）或 bulk_update 函数将所有旧名称转换为新名称。有关更多信息，请
Note: Be careful when changing names of parent or child models as this will affect historical records using the old model name. To fix this use django's update() or bulk_update function to convert all old names to new names. More info is here
选项2- django-polymorphic
OPTION 2 - django-polymorphic
使用名为的django软件包django-polymorphic ，这允许在查询父类时返回所有具体的类。
Using a django package called django-polymorphic, this allows all concrete classes to be returned when parent class is queried on.
models.py
from polymorphic.models import PolymorphicModel
class Employee(PolymorphicModel):
EMPLOYEE_TYPE_CHOICES = (
('Janitor', 'Janitor'),
('Software Engineer', 'Software Engineer'),
)
name = models.CharField(max_length=50)
email = models.CharField(max_length=80)
class Janitor(Employee):
[...]
class SoftwareEngineer(Employee):
predominant_programming_language= models.CharField(max_length=100)
[...]
在<$ c $上查询时c>员工会返回以下模型
>>> Employee.objects.filter(id=1)
[ <Employee: id 1, name "Joe Bloggs", email "[email protected]">,
<SoftwareEngineer: id 1, name "Joe Bloggs", email "[email protected]", predominant_programming_language "Python">,]
这篇关于外键中的许多型号的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！