使用django generic CreateView我可以创建一个新的用户帐户,但是如何在注册后使用这种技术自动登录该用户?
urls.py
...
url( r'^signup/$', SignUpView.as_view(), name = 'user_signup' ),
...
views.py
class SignUpView ( CreateView ) :
form_class = AccountCreationForm
template_name = 'accounts/signup.html'
success_url = reverse_lazy( 'home' )
forms.py
class AccountCreationForm ( forms.ModelForm ) :
def __init__( self, *args, **kwargs ) :
super( AccountCreationForm, self ).__init__( *args, **kwargs )
for field in self.fields :
self.fields[field].widget.attrs['class'] = 'form-control'
password1 = forms.CharField( label = 'Password', widget = forms.PasswordInput )
password2 = forms.CharField( label = 'Password confirmation', widget = forms.PasswordInput )
class Meta :
model = User
fields = ( 'email', 'first_name', )
def clean_password2 ( self ) :
# Check that the two password entries match
password1 = self.cleaned_data.get( "password1" )
password2 = self.cleaned_data.get( "password2" )
if password1 and password2 and password1 != password2:
raise forms.ValidationError( "Passwords don't match" )
return password
def save( self, commit = True ) :
# Save the provided password in hashed format
user = super( AccountCreationForm, self ).save( commit = False )
user.set_password( self.cleaned_data[ "password1" ] )
if commit:
user.save()
return user
答案 0 :(得分:15)
它可能晚了,但这正是我的问题,经过几个小时的挣扎终于找到了。
也许你找到了,但如果其他人正在寻找解决方案,那么这就是我的。
您只需在类继承CreateView中覆盖form_valid()
。以下是我自己的类的示例:
class CreateArtistView(CreateView):
template_name = 'register.html'
form_class = CreateArtistForm
success_url = '/'
def form_valid(self, form):
valid = super(CreateArtistView, self).form_valid(form)
username, password = form.cleaned_data.get('username'), form.cleaned_data.get('password1')
new_user = authenticate(username=username, password=password)
login(self.request, new_user)
return valid
我首先在form_valid()
中捕获父类方法valid
的值,因为当你调用它时,它调用form.save(),它在数据库中注册你的用户并填充你的{ {1}}创建了用户。
之后我的身份验证问题很长,返回self.object
。这是因为我使用django哈希密码调用None
,并再次对其进行身份验证。
解释这一点,以便您了解我使用authenticate()
而不是form.cleaned_data.get('username')
的原因。
我希望它对你或其他人有所帮助,因为我没有在网上找到明确的答案。
答案 1 :(得分:0)
在Django 2.2中,我没有按Bestasttung的要求进行工作。但是我在form_valid方法上做了一点改动。
class CreateAccountView(CreateView):
template_name = 'auth/create_account.html'
form_class = SignInForm
success_url = '/'
def form_valid(self, form):
valid = super().form_valid(form)
# Login the user
login(self.request, self.object)
return valid