Logowanie w Django 1.5

Question

używam modelu myuser w Django 1.5 z poczty e-mail logowania: model:

class MyUser(AbstractBaseUser): 
    email = models.EmailField(
     verbose_name='email address', 
     max_length=255, 
     unique=True, 
     db_index=True, 
    ) 
    last_name=models.CharField(max_length=30) 
    first_name=models.CharField(max_length=30) 
    second_name=models.CharField(max_length=30, blank=True) 
    post=models.CharField(max_length=30, blank=True) 

    objects = MyUserManager() 

    USERNAME_FIELD = 'email' 
    REQUIRED_FIELDS = ['last_name','first_name','second_name','post', ....] 

    def get_full_name(self): 
     return self.email 

    def get_short_name(self): 
     return self.email 

    def __unicode__(self): 
     return self.email 

    def has_perm(self, perm, obj=None): 
     return True 

    def has_module_perms(self, app_label): 
     return True 

    @property 
    def is_staff(self): 
     return self.is_admin 

Próbowałem metody z importu z django.contrib.auth.views logowania:

adresy URL :

(r'^login/$', login, {'template_name': 'enter.html'}), 

widok: nic.

Działa dla superużytkownika z klasycznego modelu użytkownika, a nie dla MyUser.

Potem próbowałem:

widok:

def login_view(request): 
    username = request.POST['email'] 
    password = request.POST['password'] 
    user = authenticate(username=username, password=password) 
    if user is not None and user.is_active: 
     login(request, user) 
     return HttpResponseRedirect("/n1.html")# Redirect to a success page. 
    else: 
     return HttpResponseRedirect("/account/invalid/")# Return a 'disabled account' error message 

szablon:

{% if form.errors %} 
<p>Something is wrong</p> 
{% endif %} 

<form action="" method="post"> 
    {% csrf_token %} 
    <label for="email">Login:</label> 
    <input type="text" name="email" value="" id="email"/> 
    <label for="password">Password:</label> 
    <input type="password" name="password" value="" id="username"> 

    <input type="submit" value="login" /> 
    <input type="hidden" name="next" value="{{next|escape}}" /> 

</form> 

URL:

(r'^login/$', login_view, {'template_name': 'enter.html'}), 

Ale mam błąd login_view() got an unexpected keyword argument 'template_name' Co robię źle?

Answer 1

login_view() got an unexpected keyword argument 'template_name' oznacza, że ​​funkcja widoku powinny mieć template_name parametr:

def login_view(request, template_name): 
    'your code' 

Jeśli nie jest to potrzebne, nie przechodzą go w urls.py:

(r'^login/$', login_view), 

UPD.

Twoja login_view obsługuje twoją metodę POST. Można go ponownie zapisać w ten sposób, aby renderować formularz na GET

def login_view(request): 
    if request.method == 'POST': 
     username = request.POST['email'] 
     password = request.POST['password'] 
     user = authenticate(username=username, password=password) 
     if user is not None and user.is_active: 
      login(request, user) 
      return HttpResponseRedirect("/n1.html") 
     return HttpResponseRedirect("/account/invalid/") 
    form = LoginForm() 
    return render(request, 'enter.html', {'login_form': LoginForm})