状态模式 - Django 模型

Question

我目前正在尝试在 Django 中实现状态模式。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

现在我怎样才能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

如果状态只能是“打开”和“关闭”两个，不要为状态创建模型，您可以在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列表和IntegerField 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)

如果你真的需要单独的状态模型，当然可以，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()