Python “QueryDict”对象没有属性“association”
已登录的管理员希望注册与管理员具有相同关联名称的成员,但会引发此错误 有什么我遗漏的吗 还是个新手,谢谢你们的帮助,伙计们 admin\models.pyPython “QueryDict”对象没有属性“association”,python,django,python-2.7,python-3.x,django-views,Python,Django,Python 2.7,Python 3.x,Django Views,已登录的管理员希望注册与管理员具有相同关联名称的成员,但会引发此错误 有什么我遗漏的吗 还是个新手,谢谢你们的帮助,伙计们 admin\models.py class Administrator(AbstractUser): ... association= models.ForeignKey(Association) class Meta: db_table = 'Administrator' from pl.admin.models import
class Administrator(AbstractUser):
...
association= models.ForeignKey(Association)
class Meta:
db_table = 'Administrator'
from pl.admin.models import Administrator
class Association(models.Model):
asoc_name = models.CharField(max_length=100)
class Meta:
db_table = 'Association'
def __str__(self):
return self.asoc_name
member\models.py
class Administrator(AbstractUser):
...
association= models.ForeignKey(Association)
class Meta:
db_table = 'Administrator'
from pl.admin.models import Administrator
class Association(models.Model):
asoc_name = models.CharField(max_length=100)
class Meta:
db_table = 'Association'
def __str__(self):
return self.asoc_name
forms.py
class RegForm(forms.ModelForm):
...
association = forms.ModelChoiceField(queryset=Association.objects.none())
...
class Meta:
model = Administrator
fields = [..., 'association', ...]
def __init__(self, user, *args, **kwargs):
super(RegForm, self).__init__(*args, **kwargs)
self.fields['association'].queryset = Association.objects.filter(
asoc_name=user.association)
views.py
def member_signup(request):
if request.method == 'POST':
form = RegForm(request.POST, request.user)
if not form.is_valid():
return render(request, 'member/member_signup.html',
{'form': form})
else:
...
asoc = form.cleaned_data.get('association')
...
Member.objects.create(...
association=asoc,
...)
user = authenticate(...
association=asoc,
...)
return redirect('/')
else:
return render(request, 'member/member_signup.html',
{'form': RegForm(request.user)})
回溯编辑
user是表单的第一个参数,但在if子句中,在request.POST之后,将其作为第二个参数传递
通常应避免更改表单实例化的签名;相反,您应该将用户作为关键字参数传递,并从kwargs处获取它。您可以提供回溯吗?@JensAstrup添加了回溯