Django:无法将关键字''解析为字段.选择是:

Question

我正在访问这个奇怪的问题ManyToManyField.

我有以下型号.

class Link(models.Model):
    title = models.CharField(max_length = 200)
    url = models.URLField(unique = True)
    tags = models.ManyToManyField(Tag)
    creation_date = models.DateTimeField(auto_now_add = True)
    user = models.ForeignKey(User)
    likes = models.ManyToManyField(User, related_name = "%(app_label)s_%(class)s_user_likes")
    dis_likes = models.ManyToManyField(User, related_name = "%(app_label)s_%(class)s_user_dis_likes")

    class Meta:
        abstract = True

class URL(Link):
    preview_image = models.URLField()
    preview_heading = models.CharField(max_length = 100)
    preview_content = models.CharField(max_length = 100)

当我尝试访问时URL.objects.get(pk=1).likes.all(),我收到Cannot resolve keyword '' into field. Choices are:...错误.

URL.objects.get(pk=1).tags.all(), URL.objects.get(pk=1).user和URL.objects.filter(likes=auser, pk=1)做工精细.

更新:

1. 字段likes和dis_likes使用south通过添加schemamigration
2. 以前我使用Django 1.6.1,更新到Django 1.6.2,问题仍然存在
3. 截断数据库,同步它以获得新表,问题仍然存在

4. 部分追溯:

   File "F:\system\env\lib\site-packages\django\db\models\manager.py" in all
     133.         return self.get_queryset()
   File "F:\system\env\lib\site-packages\django\db\models\fields\related.py" in get_queryset
     549.                 return super(ManyRelatedManager, self).get_queryset().using(db)._next_is_sticky().filter(**self.core_filters)
   File "F:\system\env\lib\site-packages\django\db\models\query.py" in filter
     590.         return self._filter_or_exclude(False, *args, **kwargs)
   File "F:\system\env\lib\site-packages\django\db\models\query.py" in _filter_or_exclude
     608.             clone.query.add_q(Q(*args, **kwargs))
   File "F:\system\env\lib\site-packages\django\db\models\sql\query.py" in add_q
     1198.         clause = self._add_q(where_part, used_aliases)
   File "F:\system\env\lib\site-packages\django\db\models\sql\query.py" in _add_q
     1234.                     current_negated=current_negated)
   File "F:\system\env\lib\site-packages\django\db\models\sql\query.py" in build_filter
     1100.                     allow_explicit_fk=True)
   File "F:\system\env\lib\site-packages\django\db\models\sql\query.py" in setup_joins
     1357.             names, opts, allow_many, allow_explicit_fk)
   File "F:\system\env\lib\site-packages\django\db\models\sql\query.py" in names_to_path
     1277.                                      "Choices are: %s" % (name, ", ".join(available)))
   
   Exception Type: FieldError at /url/3
   Exception Value: Cannot resolve keyword '' into field. Choices are: __app___article_user_dis_likes, __app___article_user_likes, __app___imageurl_user_dis_likes, __app___imageurl_user_likes, __app___review_user_dis_likes, __app___review_user_likes, __app___url_user_dis_likes, __app___url_user_likes, __app___videourl_user_dis_likes, __app___videourl_user_likes, article, date_joined, email, first_name, groups, id, imageurl, is_active, is_staff, is_superuser, last_login, last_name, logentry, password, review, url, user_permissions, username, userobjectpermission, videourl
   

Answer 1

我想我已经找到问题了。我想问题出在我的应用程序的名称上__app__。Django 字段查找假设__（双下划线）之前的所有内容都是一个字段，在我的例子中解析为``（空字符串）。

总是很难命名默认应用程序及其所在的项目。认为__app__是更Pythonic和更聪明的解决方案。我想我应该将我的应用程序重命名为app. 希望这有效。