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.
Answers:
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
See django docs for views.login(), you supply a ‘next’ value (as a hidden field) on the input form to redirect to after a successful login.
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!
This may not be a “best practice”, but I’ve successfully used this before:
return HttpResponseRedirect(request.META.get('HTTP_REFERER','/'))
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 %}
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'}),
)
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" %}
Note: The file should contain above lines only.
To support full urls with param/values you’d need:
?next={{ request.get_full_path|urlencode }}
instead of just:
?next={{ request.path }}
You can also do this
<input type="hidden" name="text" value="{% url 'dashboard' %}" />
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.
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
See django docs for views.login(), you supply a ‘next’ value (as a hidden field) on the input form to redirect to after a successful login.
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!
This may not be a “best practice”, but I’ve successfully used this before:
return HttpResponseRedirect(request.META.get('HTTP_REFERER','/'))
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 %}
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'}),
)
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" %}
Note: The file should contain above lines only.
To support full urls with param/values you’d need:
?next={{ request.get_full_path|urlencode }}
instead of just:
?next={{ request.path }}
You can also do this
<input type="hidden" name="text" value="{% url 'dashboard' %}" />