'NoneType' 对象没有属性 'unique' 但我没有使用 'unique' 属性

Question

我在 Django 1.8 中构建模型，并且我正在使用抽象继承（我假设这是导致问题的原因）。我有抽象模型，然后我有基于这些抽象模型的模型。我也有一些模型之间的 ForeignKey 和 ManyToMany 关系。

一切看起来都很好，但是当我尝试 syncdb 或“makemigrations blog”时，我得到一个 AttributeError，上面写着“NoneType”对象没有“唯一”属性。

我不知道我为什么会得到它，我尝试了不同的模型设置，我阅读了很多论坛帖子，但现在我碰壁了。

我将在下面发布回溯和我的模型：

型号：

独立数据库

from django.db import models

class URL(models.Model):
    link = models.CharField(max_length=400)
    name = models.CharField(max_length=200)

    def __str__(self):
        return self.name


class Artist(models.Model):
    name = models.CharField(max_length=200)
    description = models.TextField(null=True, blank=True)
    link = models.ForeignKey(URL)

    class Meta:
        abstract = True
        ordering = ['name']

    def __str__(self):
        return self.name


class ArtistSingle(Artist):
    birthdate = models.DateField(null=True, blank=True)
    deathdate = models.DateField(null=True, blank=True)


class ArtistGroup(Artist):
    members = models.ManyToManyField(ArtistSingle)
    established = models.DateField(null=True, blank=True)
    disbanded = models.DateField(null=True, blank=True)


class Contributor(models.Model):
    contributing_artist = models.ForeignKey(ArtistSingle, null=True, blank=True)
    alternate_name = models.CharField(max_length=200)
    role = models.CharField(max_length=200)


class ProductionCompany(models.Model):
    name = models.CharField(max_length=200)
    link = models.ForeignKey(URL)

    def __str__(self):
        return self.name


class Work(models.Model):
    title = models.CharField(max_length=200)
    description = models.TextField(null=True, blank=True)
    created = models.DateField()
    city = models.CharField(max_length=200)
    production_company = models.ForeignKey(ProductionCompany, blank=True, null=True)
    self_published = models.BooleanField(default=False)
    creator = models.ForeignKey(Artist)
    link = models.ForeignKey(URL)
    styles = models.CharField(max_length=200)
    contributors = models.ManyToManyField(Contributor)

    class Meta:
        abstract = True
        ordering = ['-created']

    def __str__(self):
        return self.title


class MusicalWork(Work):
    audio_links = models.ManyToManyField(URL)


class WrittenWork(Work):
    excerpt = models.TextField(null=True, blank=True)


class PerformanceWork(Work):
    venue = models.CharField(max_length=200)


class VideoWork(Work):
    length = models.CharField(max_length=16)


class VisualWork(Work):
    images = models.ManyToManyField(URL)

博客：

from django.db import models

# Create your models here.


class Tag(models.Model):
    tag_name = models.CharField(max_length=200)

    def __str__(self):
        return self.tag_name


class Entry(models.Model):
    title = models.CharField(max_length=200)
    body = models.TextField()
    slug = models.SlugField(max_length=200)
    publish = models.BooleanField(default=True)
    created = models.DateTimeField(auto_now=True)
    modified = models.DateTimeField(auto_now=True)
    tags = models.ManyToManyField(Tag)

    def __str__(self):
        return self.title

    class Meta:
        verbose_name = "Blog Entry"
        verbose_name_plural = "Blog Entries"
        ordering = ["-created"]

回溯：

[pattmayne@web476 weird_canada]$ python3.4 manage.py makemigrations blog
Traceback (most recent call last):
  File "manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/management/__init__.py", line 338, in execute_from_command_line
    utility.execute()
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/management/__init__.py", line 330, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/management/base.py", line 390, in run_from_argv
    self.execute(*args, **cmd_options)
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/management/base.py", line 440, in execute
    self.check()
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/management/base.py", line 478, in check
    include_deployment_checks=include_deployment_checks,
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/checks/registry.py", line 72, in run_checks
    new_errors = check(app_configs=app_configs)
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/core/checks/model_checks.py", line 28, in check_all_models
    errors.extend(model.check(**kwargs))
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/db/models/base.py", line 1181, in check
    errors.extend(cls._check_fields(**kwargs))
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/db/models/base.py", line 1258, in _check_fields
    errors.extend(field.check(**kwargs))
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/db/models/fields/related.py", line 1829, in check
    errors = super(ForeignKey, self).check(**kwargs)
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/db/models/fields/related.py", line 1502, in check
    errors.extend(self._check_unique_target())
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/db/models/fields/related.py", line 1522, in _check_unique_target
    for rel_field in self.foreign_related_fields)
  File "/home/pattmayne/webapps/limbs_008/lib/python3.4/Django-1.8.2-py3.4.egg/django/db/models/fields/related.py", line 1522, in <genexpr>
    for rel_field in self.foreign_related_fields)
AttributeError: 'NoneType' object has no attribute 'unique'

如果这是由我继承的模型引起的，确切的原因是什么，我应该如何改变？

提前致谢。

Answer 1

我认为问题实际上出在 Work 模型上。

您在抽象类中有一个指向 URL 的 ForeignKey，

link = models.ForeignKey(URL)

而且在某些派生类中，您还拥有URL 的密钥，例如 MusicalWork：

class MusicalWork(Work):
    audio_links = models.ManyToManyField(URL)

所以 MusicalWork 有两个指向 URL 的链接。这很好，除了 Django 尝试为你的模型的 URL 创建一个反向关系，在这种情况下通常称为musicalwork_set，但它对于同一个模型有两个反向关系！

答案是为任何从它派生的具有相同模型引用的模型指定一个related_name 字段。

因此：

class MusicalWork(Work):
    audio_links = models.ManyToManyField(URL, related_name='musicalwork_audio_set')

但可能还有其他问题，因为该错误消息并不完全指向这种情况（相信我，对于这个特定错误，Django 有更好的错误消息）。

Answer 2

这是因为您有一个抽象模型的外键。

creator = models.ForeignKey(Artist)

你可以把这个改成ArtistSingle

creator = models.ForeignKey(ArtistSingle)

理想情况下它应该抛出一个适当的错误，例如

字段定义了与模型“艺术家”的关系，该模型要么未安装，要么是抽象的。

但崩溃是bug with Django 1.8