python - 在 Django 模型中的定义之前使用一个类

Question

当我尝试 syncdb 时，出现错误 Menu is not a valid class Name。

我怎样才能解决这个关系案例:

class MenuItem(model.Models)
    title = models.CharField(max_length=200)
    submenus = models.ManyToManyField(Menu, blank=True, null=True)

class Menu(Container):
    links = models.ManyToManyField(MenuItem)

Answer 1

来自 the Django book :

If you need to create a relationship on a model that has not yet been defined, you can use the name of the model, rather than the model object itself:

例如:

class MenuItem(model.Models)
    title = models.CharField(max_length=200)
    submenus = models.ManyToManyField('Menu', blank=True, null=True)
                                      ^    ^

编辑:
作为Francis mentions (如 the documentation 中所写):

It doesn't matter which model has the ManyToManyField, but you should only put it in one of the models -- not both.