Question

如何继承ModelManager？

class Content(models.Model):
    name = models.CharField(max_length=255, verbose_name='Name des Blogs')
    slug = models.SlugField(max_length=80, blank=True)
    objects = models.Manager()
    active = ContentActiveManager()

class ContentActiveManager(models.Manager):
    def get_query_set(self):
        return super(ContentActiveManager,self).get_query_set().filter(activated=True,show=True)

class BlogCatalog(Content):
    frequency = models.PositiveSmallIntegerField(max_length=2, choices=make_index_based_tuple(l=FREQUENCY), verbose_name='Frequenz',)



blog = BlogCatalog.active.get(pk=1)

blog现在显然是一个Content对象。如果我输入Catalog.active.get（pk = 1）我想要一个Content对象但是如果我键入BlogCatalog.active.get（pk = 1），我想要一个BlogCatalog对象。

如何在不浪费的情况下实现这一目标？

Answer 1

Django只允许从抽象基类继承Manager。要将同一个经理用作非ABC，您必须明确声明它。

查看django docs on custom managers and inheritance。

基本上，只需这样做：

class BlogCatalog(Content):
    frequency = models.PositiveSmallIntegerField(max_length=2, choices=make_index_based_tuple(l=FREQUENCY), verbose_name='Frequenz',)
    active = ContentActiveManager()

希望有所帮助。

Answer 2

我想出的唯一方法：

class Content:

    @staticmethod
    def __new__(cls, *args, **kwargs):
        super_new = super(Content, cls).__new__(cls, *args, **kwargs)
        cls.add_to_class('active', ContentActiveManager())
        return super_new

django ModelManager继承

2 个答案: