状态模式- Django模型

Question

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，我希望能够这样做：

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'open‘和'closed'，你可以在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()

您还可以在模型中使用预定义的状态和IntegerField列表定义状态：

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)

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()