python - 登录 Django 1.5

Question

我在 Django 1.5 中使用 MyUser 模型和电子邮件登录:型号:

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

我尝试了使用 from django.contrib.auth.views import login 的方法:

网址:

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

View :无。

它适用于经典用户模型中的 super 用户，不适用于 MyUser。

然后我尝试了:

查看:

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

模板:

{% 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>

网址:

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

但是我得到了错误 login_view() got an unexpected keyword argument 'template_name'我做错了什么？

Answer 1

login_view() got an unexpected keyword argument 'template_name' 意味着你的 View 函数应该有 template_name 参数:

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

如果你不需要它，不要在urls.py中传递它:

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

更新。

您的 login_view 处理您的 POST 方法。您可以用这种方式重写它以在 GET 上呈现表单

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})