Question

我正在尝试构建一个管理面板，用户可以使用某些规则填充数据库;对于某些实体/模型，每个用户都可以查看和编辑自己的数据。

为此，我以这种方式继承了一个modelAdmin类

#my models

class Product(models.Model):
    name = models.CharField(max_length=80)
    description = models.TextField()
    author = models.ForeignKey(User)

    def __unicode__(self):
        return self.name

class Variant(models.Model):
    size = models.DecimalField(max_digits=3, decimal_places=0)
    author = models.ForeignKey(User)
    super_product = models.ForeignKey(Product)

    def __unicode__(self):
        return "%s %s" % (self.size)

#in urls.py

class FilterProduct(admin.ModelAdmin):
    def queryset(self, request):
        qs = super(FilterProduct, self).queryset(request)
        if request.user.is_superuser:
            return qs
        return qs.filter(author=request.user)

    def save_model(self, request, obj, form, change):
        if not request.user.is_superuser:
            obj.author = request.user

        obj.save()

class FilterVariant(admin.ModelAdmin):
    def queryset(self, request):
        qs = super(FilterVariant, self).queryset(request)
        if request.user.is_superuser:
            return qs
        return qs.filter(author=request.user)

    def save_model(self, request, obj, form, change):
        if not request.user.is_superuser:
            obj.author = request.user

        obj.save()

    def get_form(self, request, obj, **kwargs):
        form = super(FilerVariant,self).get_form(self,request, obj,**kwargs)

        if not request.user.is_superuser:
            form.base_fields['super_product'].queryset = form.base_fields['super_product'].queryset.filter(author=request.user)

        return form

admin.site.register(Product,FilterProduct)
admin.site.register(Variant,FilterVariant)

当我尝试添加Variant时，我收到此错误

Environment:

Request Method: GET
Request URL: http://127.0.0.1:8000/admin/prodotti/varianti/add/
Django Version: 1.2.3
Python Version: 2.6.6
Installed Applications:
['django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.sites',
 'django.contrib.messages',
 'prodotti',
 'django.contrib.admin']
Installed Middleware:
('django.middleware.common.CommonMiddleware',
 'django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware')


Traceback:
File "/usr/lib/pymodules/python2.6/django/core/handlers/base.py" in get_response
  100.                     response = callback(request, *callback_args, **callback_kwargs)
File "/usr/lib/pymodules/python2.6/django/contrib/admin/options.py" in wrapper
  265.                 return self.admin_site.admin_view(view)(*args, **kwargs)
File "/usr/lib/pymodules/python2.6/django/utils/decorators.py" in _wrapped_view
  76.                     response = view_func(request, *args, **kwargs)
File "/usr/lib/pymodules/python2.6/django/views/decorators/cache.py" in _wrapped_view_func
  69.         response = view_func(request, *args, **kwargs)
File "/usr/lib/pymodules/python2.6/django/contrib/admin/sites.py" in inner
  190.             return view(request, *args, **kwargs)
File "/usr/lib/pymodules/python2.6/django/utils/decorators.py" in _wrapper
  21.             return decorator(bound_func)(*args, **kwargs)
File "/usr/lib/pymodules/python2.6/django/utils/decorators.py" in _wrapped_view
  76.                     response = view_func(request, *args, **kwargs)
File "/usr/lib/pymodules/python2.6/django/utils/decorators.py" in bound_func
  17.                 return func(self, *args2, **kwargs2)
File "/usr/lib/pymodules/python2.6/django/db/transaction.py" in _commit_on_success
  299.                     res = func(*args, **kw)
File "/usr/lib/pymodules/python2.6/django/contrib/admin/options.py" in add_view
  799.         ModelForm = self.get_form(request)

Exception Type: TypeError at /admin/prodotti/varianti/add/
Exception Value: get_form() takes exactly 3 arguments (2 given)

也许是一个愚蠢的问题，错误在哪里？我给函数提供了所有需要的参数吗？

Answer 1

这里有一些错误。

首先，get_form方法的原始签名是def get_form(self, request, obj=None, **kwargs) - 也就是说，obj参数是可选的（这是有道理的，就像您创建新项目时一样）没有现有的对象）。但是，你已经用它覆盖了它：def get_form(self, request, obj, **kwargs) - 即现在需要obj参数。

除非你完全控制你的方法将被调用 - 在这种情况下你没有完全控制，因为它是由管理员完成的 - 你应该确保你的方法可以接受与原始方法相同的参数，在至少。

您的第二个错误出现在下一行：

form = super(FilerVariant,self).get_form(self,request, obj,**kwargs)

这里你复制了self参数 - 你不能在方法调用中显式传递它，因为它已作为第一个参数传递。

最后，你一定要考虑升级 - 两年前发布了Django 1.2，从那以后发生了很多变化。

django 1.2 modelAdmin继承了类，错误“get_form（）正好需要3个参数（给定2个）”

1 个答案: