当debug在引发表单错误时为false时，如何使Django表单方法变为not 500？

我有以下登录代码：

class LoginForm(forms.Form):
    email = forms.EmailField(max_length = 254, min_length = 6)
    password = forms.CharField(min_length = 8, widget = forms.PasswordInput())

    def Login(self):
        email = self.cleaned_data.get('email')
        password = self.cleaned_data.get('password')
        user = authenticate(email = email, password = password)

        if (user is None) or (user is not None and user.is_active is False):
            raise forms.ValidationError('Login is incorrect.')

        return user

然而，当我在我的站点上把debug设置为false时，只要登录不正确，我就会得到一个500错误，而不是显示ValidationError的表单。我这样称呼它：

if form.is_valid():
    user = form.Login()

这是一个视图方法。我错过了什么？如何正确调用此表单错误，以便在debug设置为false时不会出错？

这样调用

def clean(self):
    password = self.cleaned_data.get('password')
    email = self.cleaned_data.get('email')   
    user = authenticate(email = email, password = password)
    if (user is None) or (user is not None and user.is_active is False):
        raise forms.ValidationError('Login is incorrect.')
    return self.cleaned_data

def login(self,request):
    user = authenticate(email = self.cleaned_data.get('email'),password = self.cleaned_data.get('password'))
    user = authenticate(username=username, password=password)
    return user

---

是否应该调用

Login

而不调用

def clean

？Hi可能重复，您是对的，我的解决方案是