django-taggit 在自定义标签上调用名称（）时出错

Question

我正在尝试创建一个模型，其中包含与公司和主题相关联的两种类型的标签，并遵循 django-taggit 中的文档制作自定义标签，以便在一个模型中拥有两个 Taggable Manager。

我的模型.py

from django.db import models
from taggit.managers import TaggableManager
from taggit.models import GenericTaggedItemBase, TagBase

# Create your models here.
class TopicTag(TagBase):
    class Meta:
        verbose_name = "TopicTag"
        verbose_name_plural = "TopicTags"

class CompanyTag(TagBase):
    class Meta:
        verbose_name = "CompanyTag"
        verbose_name_plural = "CompanyTags"

class ThroughTopicTag(GenericTaggedItemBase):
    tag = models.ForeignKey(TopicTag, on_delete=models.CASCADE)

class ThroughCompanyTag(GenericTaggedItemBase):
    tag = models.ForeignKey(CompanyTag, on_delete= models.CASCADE)


class Questions(models.Model):
    name = models.CharField(max_length=255)
    path = models.CharField(max_length=500)
    company_tags = TaggableManager(blank = True, through=ThroughCompanyTag, related_name="company_tags")
    topic_tags = TaggableManager(blank = True, through=ThroughTopicTag, related_name="topic_tags")

现在，当我尝试在 django shell 中运行以下命令时

from QuestionBank.models import Questions
Questions.objects.first().company_tags.names()

我收到以下错误：

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/taggit/utils.py", line 124, in inner
    return func(self, *args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/taggit/managers.py", line 248, in names
    return self.get_queryset().values_list("name", flat=True)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/taggit/managers.py", line 74, in get_queryset
    return self.through.tags_for(self.model, self.instance, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/taggit/models.py", line 155, in tags_for
    return cls.tag_model().objects.filter(**kwargs).distinct()
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/manager.py", line 82, in manager_method
    return getattr(self.get_queryset(), name)(*args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 904, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 923, in _filter_or_exclude
    clone.query.add_q(Q(*args, **kwargs))
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1350, in add_q
    clause, _ = self._add_q(q_object, self.used_aliases)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1377, in _add_q
    child_clause, needed_inner = self.build_filter(
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1250, in build_filter
    lookups, parts, reffed_expression = self.solve_lookup_type(arg)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1087, in solve_lookup_type
    _, field, _, lookup_parts = self.names_to_path(lookup_splitted, self.get_meta())
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1482, in names_to_path
    raise FieldError("Cannot resolve keyword '%s' into field. "
django.core.exceptions.FieldError: Cannot resolve keyword 'None' into field. Choices are: company_tags, id, name, slug, throughcompanytag

提前致谢！

Answer 1

我认为下面的代码可以解决您的问题：

[t.name for t in your_tags.all()]

我在我的models.py定义了这个，下面的例子可能有帮助：

class TagsAdmin(models.Model):
    admin_tags = TaggableManager()
    admin = models.ForeignKey(AdminInit, on_delete=models.CASCADE)

    class Meta:
        verbose_name = _("TagsAdmin")
        ordering = ("created",)

    def __str__(self):
        return '{0}'.format([t.name for t in self.admin_tags.all()])