python - 带有 list_filter 错误 : FieldDoesNotExist: Region has no field named 'p' 的 Wagtail ModelAdmin

Question

我有一个通过 Wagtail 的 modeladmin 公开的模型，在我尝试应用包含 ForeignKey 的 list_filter 之前一直有效。我有一个名为 Region 的自引用模型(用于存储县、市和镇)，我希望能够按 Parent 级别在其上创建过滤器。

我在 Django 1.10 上运行 Wagtail 1.8.1。这是模型:

class Region(models.Model):
    """
    Tree of regions and sub-regions.
    """

    name = models.CharField(max_length=255)
    parent = models.ForeignKey(
        'Region',
        blank=True,
        null=True,
    )

    class Meta:
        unique_together = ('name', 'parent')

    def __str__(self):
        return '{0}'.format(
            self.name,
        )

wagtail_hooks.py 中的 ModelAdmin:

class RegionAdmin(ModelAdmin):
    model = Region
    menu_icon = 'doc-full-inverse'
    empty_value_display = 'ROOT'
    list_display = ('parent', 'name')
    list_filter = ('parent')


modeladmin_register(RegionAdmin)

如果我注释掉 RegionAdmin 类中的 list_filter 行，它会正常工作。文档说它可以在列表中使用 ForeignKey。我错过了一些明显的东西吗？堆栈跟踪在这里:

Traceback (most recent call last):
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/core/handlers/exception.py", line 39, in inner
    response = get_response(request)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/core/handlers/base.py", line 187, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/core/handlers/base.py", line 185, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/views/decorators/cache.py", line 43, in _cache_controlled
    response = viewfunc(request, *args, **kw)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/wagtail/wagtailadmin/decorators.py", line 24, in decorated_view
    return view_func(request, *args, **kwargs)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/wagtail/contrib/modeladmin/options.py", line 350, in index_view
    return view_class.as_view(**kwargs)(request)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/views/generic/base.py", line 68, in view
    return self.dispatch(request, *args, **kwargs)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/utils/decorators.py", line 67, in _wrapper
    return bound_func(*args, **kwargs)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/contrib/auth/decorators.py", line 23, in _wrapped_view
    return view_func(request, *args, **kwargs)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/utils/decorators.py", line 63, in bound_func
    return func.__get__(self, type(self))(*args2, **kwargs2)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/wagtail/contrib/modeladmin/views.py", line 249, in dispatch
    self.queryset = self.get_queryset(request)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/wagtail/contrib/modeladmin/views.py", line 554, in get_queryset
    filters_use_distinct) = self.get_filters(request)
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/wagtail/contrib/modeladmin/views.py", line 381, in get_filters
    field_path)[-1]
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/contrib/admin/utils.py", line 485, in get_fields_from_path
    fields.append(parent._meta.get_field(piece))
  File "/home/vagrant/.virtualenvs/sepia/lib/python3.5/site-packages/django/db/models/options.py", line 619, in get_field
    raise FieldDoesNotExist('%s has no field named %r' % (self.object_name, field_name))
django.core.exceptions.FieldDoesNotExist: Region has no field named 'p'

感谢任何帮助或想法!

Answer 1

你少了一个逗号:

list_filter = ('parent',)

如果没有逗号，括号将被忽略，django 将迭代字符串，而不是元组(因为没有元组!)，因此尝试访问字段 p。

来自docs :

a tuple with one item is constructed by following a value with a comma (it is not sufficient to enclose a single value in parentheses). Ugly, but effective.