相关疑难解决方法(0)

似乎有很多方法可以在Python中定义单例.Stack Overflow是否有共识？

一个例子胜过千言万语:

   In [3]: User.objects.filter(id=19)[0] == User.objects.filter(id=19)[0]
   Out[3]: True

   In [4]: User.objects.filter(id=19)[0] == User.objects.filter(id=19).defer('email')[0]
   Out[4]: False

它故意这样工作吗？

子问题:是否有任何简单的方法可以从延迟模型中获取常规模型实例？

编辑:

看起来contenttypes框架已正确修补:http://code.djangoproject.com/changeset/10523

所以我会说Model ._____ eq _____()运算符不应该是这样的:

    def __eq__(self, other):
        return isinstance(other, self.__class__) and self._get_pk_val() == other._get_pk_val()

但更像这样:

    def __eq__(self, other):
        return ContentType.objects.get_for_model(self) is ContentType.objects.get_for_model(other) and self._get_pk_val() == other._get_pk_val()

这当然首次导致两次DB命中,但幸运的是get_for_model似乎实现了缓存.

我有一堂课

class PlaylistManager(models.Manager):
    def add_playlist(self, name):
        playlist = Playlist(name=name)
        playlist.save()
        return playlist

    def get_playlist_with_id(self, id):
        return super(PlaylistManager, self).get_query_set().filter(pk=id)

class Playlist(models.Model):
    name = models.CharField(max_length=30)
    date_created = models.DateTimeField(auto_now_add=True)
    date_modified = models.DateTimeField(auto_now=True)
    deleted = models.BooleanField(default=False)
    objects = PlaylistManager() # is a customer manager

    def __repr__(self):
        return '<Playlist name:%s, date_created:%s, date_modified:%s, deleted:%s>' % \
                (self.name, self.date_created, self.date_modified, self.deleted)

    class Meta:
        db_table = 'playlists'

和我test一样

def test_get_playlist(self):
    playlist = Utility.add_playlist()
    self.assertEqual(Playlist.objects.get_playlist_with_id(playlist.id), playlist)

class Utility():
    @staticmethod
    def add_playlist(playlist_name=PLAYLIST):
        return Playlist.objects.add_playlist(playlist_name)

当我运行测试时,我发现错误为

AssertionError: [<Playlist name:playlist, …