Python 如何从表单中获取用户名?
我今天遇到了一个特殊的情况。以前我在view.py中有以下内容Python 如何从表单中获取用户名?,python,django,forms,Python,Django,Forms,我今天遇到了一个特殊的情况。以前我在view.py中有以下内容 def register_page(request): if request.method == 'POST': form = RegistrationForm(request.POST) if form.is_valid(): user = User.objects.create_user( username=form.cleaned_
def register_page(request):
if request.method == 'POST':
form = RegistrationForm(request.POST)
if form.is_valid():
user = User.objects.create_user(
username=form.cleaned_data['username'],
password=form.cleaned_data['password2'],
email=form.cleaned_data['email']
)
return HttpResponseRedirect('/register/success/')
else:
form = RegistrationForm()
variables = RequestContext(request, {'form':form})
return render_to_response('registration/register.html', variables)
在她注册后,检索用户名、电子邮件和密码来创建一个新用户非常简单。但现在我已经对它进行了重构,使用哈希代码作为用户名,并单独使用电子邮件进行注册和登录
缩短的注册表单如下所示:
class RegistrationForm(forms.ModelForm):
email = forms.EmailField(label=_("Email"))
password1 = forms.CharField(label=_("Password"), widget=forms.PasswordInput)
password2 = forms.CharField(label=_("Password confirmation"), widget=forms.PasswordInput))
class Meta:
model = User
fields = ("email",)
def save(self, commit=True):
user = super(RegistrationForm, self).save(commit=False)
user.set_password(self.cleaned_data["password1"])
email = self.cleaned_data['email']
user.username = md5(email).digest().encode('base64')[:-1]
if commit:
user.save()
return user
class cb_user_model_backend(ModelBackend):
def create_user(self, email=None, password=None):
"""
Creates and saves a User with the given email and password only.
"""
now = timezone.now()
username = md5(email).digest().encode('base64')[:-1]
email = UserManager.normalize_email(email)
user = self.model(username=username, email=email,
is_staff=False, is_active=True, is_superuser=False,
last_login=now, date_joined=now)
user.set_password(password)
user.save(using=self._db)
return user
新表单不再具有用户名,因为它是经过计算的,不再由用户输入。但是如何从视图中检索用户名?新代码不是我写的,我在博客上看到的。也许钥匙在这里的元类?从文档中,我无法完全理解他试图通过这里的Meta类实现什么
非常感谢,
编辑:
好的,我想我现在明白了子类应该如何工作了。我尝试将User类分为以下子类:
class RegistrationForm(forms.ModelForm):
email = forms.EmailField(label=_("Email"))
password1 = forms.CharField(label=_("Password"), widget=forms.PasswordInput)
password2 = forms.CharField(label=_("Password confirmation"), widget=forms.PasswordInput))
class Meta:
model = User
fields = ("email",)
def save(self, commit=True):
user = super(RegistrationForm, self).save(commit=False)
user.set_password(self.cleaned_data["password1"])
email = self.cleaned_data['email']
user.username = md5(email).digest().encode('base64')[:-1]
if commit:
user.save()
return user
class cb_user_model_backend(ModelBackend):
def create_user(self, email=None, password=None):
"""
Creates and saves a User with the given email and password only.
"""
now = timezone.now()
username = md5(email).digest().encode('base64')[:-1]
email = UserManager.normalize_email(email)
user = self.model(username=username, email=email,
is_staff=False, is_active=True, is_superuser=False,
last_login=now, date_joined=now)
user.set_password(password)
user.save(using=self._db)
return user
我现在面临的问题是两个错误,self.\u db和self.model,它们应该位于基本用户类上。你从这里怎么去
编辑2:
PyCharm抱怨两个self._db和seld.model在当前cb_user_model_后端上不存在
请注意,该视图经过重构以采用两个参数:
user = User.objects.create_user(
password=form.cleaned_data['password2'],
email=form.cleaned_data['email']
)
运行时,堆栈跟踪为:
Exception Type: TypeError
Exception Value:
create_user() takes at least 2 arguments (3 given)
尝试在models.py中对save方法进行子类化:
def save(self, *args, **kwargs):
if not self.id:
self.username = md5(self.email).digest().encode('base64')[:-1]
super(ModelName, self).save(*args, **kwargs)
调用user.save()后,user.username应在视图中生成用户名。希望这有帮助
编辑:
如果要调用create_user(**kwargs),可以在views.py中执行以下操作:
email = self.cleaned_data['email']
username = md5(email).digest().encode('base64')[:-1]
u = User.objects.create_user(username = username, email = email, password = password)
但是在创建用户对象user.objects.create_user(…)时,我仍然需要一个用户名作为参数传递,否?@Kave是的,您需要。您可以按照我对save子类化的方式将create_user子类化,以自动生成一个用户名,就像使用md5一样。或者你可以做我在编辑中发布的事情。是的,你是对的,我可以把它从表单重构到视图中。但我也尝试了子类的想法,并创建了一个,有两个小问题。请参见编辑。谢谢你发布完整的堆栈跟踪,@Kave?写一条评论,说明为什么有人会否决一个有效的问题,这是公平的。非常有建设性!