Django: Redirect to previous page after login

Question

I'm trying to build a simple website with login functionality very similar to the one here on SO. The user should be able to browse the site as an anonymous user and there will be a login link on every page. When clicking on the login link the user will be taken to the login form. After a successful login the user should be taken back to the page from where he clicked the login link in the first place. I'm guessing that I have to somehow pass the url of the current page to the view that handles the login form but I can't really get it to work.

EDIT: I figured it out. I linked to the login form by passing the current page as a GET parameter and then used 'next' to redirect to that page. Thanks!

EDIT 2: My explanation did not seem to be clear so as requested here is my code: Lets say we are on a page foo.html and we are not logged in. Now we would like to have a link on foo.html that links to login.html. There we can login and are then redirected back to foo.html. The link on foo.html looks like this:

      <a href='/login/?next={{ request.path }}'>Login</a> 

Now I wrote a custom login view that looks somewhat like this:

def login_view(request):
   redirect_to = request.REQUEST.get('next', '')
   if request.method=='POST':
      #create login form...
      if valid login credentials have been entered:
         return HttpResponseRedirect(redirect_to)  
   #...
   return render_to_response('login.html', locals())

And the important line in login.html:

<form method="post" action="./?next={{ redirect_to }}">

So yeah thats pretty much it, hope that makes it clear.

Answer 1

You do not need to make an extra view for this, the functionality is already built in.

First each page with a login link needs to know the current path, and the easiest way is to add the request context preprosessor to settings.py (the 4 first are default), then the request object will be available in each request:

settings.py:

TEMPLATE_CONTEXT_PROCESSORS = (
    "django.core.context_processors.auth",
    "django.core.context_processors.debug",
    "django.core.context_processors.i18n",
    "django.core.context_processors.media",
    "django.core.context_processors.request",
)

Then add in the template you want the Login link:

base.html:

<a href="{% url django.contrib.auth.views.login %}?next={{request.path}}">Login</a>

This will add a GET argument to the login page that points back to the current page.

The login template can then be as simple as this:

registration/login.html:

{% block content %}
<form method="post" action="">
  {{form.as_p}}
<input type="submit" value="Login">
</form>
{% endblock %}

Answer 2

To support full urls with param/values you'd need:

?next={{ request.get_full_path|urlencode }}

instead of just:

?next={{ request.path }}

Answer 3

这可能不是“最佳实践”，但我之前已经成功使用过：

return HttpResponseRedirect(request.META.get('HTTP_REFERER','/'))

Answer 4

Django's built-in authentication works the way you want.

Their login pages include a next query string which is the page to return to after login.

Look at http://docs.djangoproject.com/en/dev/topics/auth/#django.contrib.auth.decorators.login_required

Answer 5

I encountered the same problem. This solution allows me to keep using the generic login view:

urlpatterns += patterns('django.views.generic.simple',
    (r'^accounts/profile/$', 'redirect_to', {'url': 'generic_account_url'}),
)

Answer 6

I linked to the login form by passing the current page as a GET parameter and then used 'next' to redirect to that page. Thanks!

Answer 7

In registration/login.html (nested within templates folder) if you insert the following line, the page will render like Django's original admin login page:

{% include "admin/login.html" %}

Answer 8

有关views.login() ，请参阅 django 文档，您在输入表单上提供“下一个”值（作为隐藏字段）以在成功登录后重定向到。

Answer 9

你也可以这样做

<input type="hidden" name="text" value="{% url 'dashboard' %}" />