Django自定义登录页面

Question

我有一个自定义的Django登录页面.我想在用户名或密码字段为空时抛出异常.我怎样才能做到这一点？

我的view.py登录方法:

def user_login(request):
    context = RequestContext(request)
    if request.method == 'POST':
        # Gather the username and password provided by the user.
        # This information is obtained from the login form.
        username = request.POST['username']
        password = request.POST['password']
        user = authenticate(username=username, password=password)
        print("auth",str(authenticate(username=username, password=password)))

        if user:
            # Is the account active? It could have been disabled.
            if user.is_active:
                login(request, user)
                return HttpResponseRedirect('/')
        else:
            return HttpResponse("xxx.")
    else:
        # Bad login details were provided. So we can't log the user in.
        print ("Invalid login details: {0}, {1}".format(username, password))
        return HttpResponse("Invalid login details supplied.")
    else:
        return render_to_response('user/profile.html', {}, context)

我试过这个并没有用:这是forms.py

def clean_username(self):
    username = self.cleaned_data.get('username')
    if not username:
        raise forms.ValidationError('username does not exist.')

Answer 1

您可以使用Django提供的登录视图.所以你的login.html应该看起来像那个例子.

<form class="login" method="POST" action="/login/">

  {% csrf_token %}

  {{form.as_p}}

  <li><input type="submit" class="logBut" value="Log in"/></li>
  </form>

并记住urls.py!

url(r'^login/$','django.contrib.auth.views.login', {'template_name': '/login.html'}),

Answer 2

正确的方法是使用表格，而不是直接从中获取变量request.POST。然后，Django将验证表单数据，并在模板中呈现表单时显示错误。如果需要一个表单字段，那么当该字段为空时，Django将自动显示错误，您甚至不需要为此编写clean_<field_name>方法。

Django已经有一个内置的登录视图。最简单的方法是使用它而不是自己编写。如果您仍然想编写自己的视图，那么查看Django的工作方式仍然很有用。