状态模式-Django模型

Question

我目前正在Django中尝试实现状态模式。以这些模型为例：

class Restaurant(models.Model):
    name = models.CharField()
    # other fields here ...

class State(models.Model):
    pass
class StateOpen(State):
    def toggle_open_closed():
        pass
class StateClosed(State):
    def toggle_open_closed():
        pass

现在，我如何让我的餐厅有一个状态，这个状态可以是StateOpen或StateClosed？

编辑：理想情况下，我希望能够做这样的事情：

r = Restaurant(name='whatever')
r.state.doSomething()
# doSomething() being a function that each state child class has,
# but implemented differently

Answer 1

不要为状态创建模型，如果状态只能是两个"打开"one_answers"关闭"，则可以在Restaurant模型中创建状态字段：

class Restaurant(models.Model):
    name = models.CharField()
    state = models.BooleanField(default=False)
    def toggle_open_closed(self):
        self.state = not self.state
        self.save()

您还可以定义状态，我们预定义的状态列表和模型中的IntegerField：

RESTARAUNT_STATE = (
    (0, 'Open'),
    (1, 'Closed'),
    (2, 'Didnt decided yet, come here later!'),
    # you can define more states later
)
class Restaurant(models.Model):
    name = models.CharField()
    state = models.IntegerField(choices=RESTARAUNT_STATE)

如果您真的需要单独的状态模型，您当然可以创建它，但toggle_state函数必须在Restaraunt模型中。

class State(models.Model):
    name_of_state = models.CharField()
class Restaurant(models.Model):
    name = models.CharField()
    state = models.ForeignKey(State)
    def toggle_state(self):
        self.state = State.objects.get(...)
        self.save()