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.
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:
Login
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 %}
To support full urls with param/values you'd need:
?next={{ request.get_full_path|urlencode }}
instead of just:
?next={{ request.path }}
This may not be a "best practice", but I've successfully used this before:
return HttpResponseRedirect(request.META.get('HTTP_REFERER','/'))
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
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!
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.
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.
You can also do this
<input type="hidden" name="text" value="{% url 'dashboard' %}" />
Related
I have seen a number of forums and posts but still couldn't get the handle of it. Here in django doc, it says
The CSRF middleware is activated by default in the MIDDLEWARE setting. If you override that setting, remember that 'django.middleware.csrf.CsrfViewMiddleware' should come before any view > middleware that assume that CSRF attacks have been dealt with.
If you disabled it, which is not recommended, you can use csrf_protect() on particular views you want to protect (see below).
In any template that uses a POST form, use the csrf_token tag inside the > element if the form is for an internal URL, e.g.:
form action
{% csrf_token %}
Based on that, in my html template I did simply:
<form id='frm' name='frm' method="post" action="{% url 'gettip' %}" >
{% csrf_token %}
<input type="text" name="tipid" name="tipid">
<input type="submit" value="Get Tip Value"/>
</form>
I expected the CSRF_token to create the hidden element since the middleware is already loaded. I see no element in the form and I get CSRF error.
The form is not associated with any model. I haven't used forms.py either. My current view is simply to output something:
def gettip(request):
if request.POST:
return HttpResponse('You requested a tip')
#from a weblink, i was told to add the following but it made no difference
context = {}
return render_to_response('tip.html',context, context_instance=RequestContext(request))
The error I am getting is obviously CSRF missing cos the hidden element is not there at all.
I am migrating from PHP and this is giving me a hard time. Though my form is not for login purposes, I couldn't get this one to work either for the same error. I am on django 1.10 and just want to get a positive response when form is submitted.
Don't use render_to_response, it's obsolete. Use render instead.
from django.shortcuts import render
def gettip(request):
if request.POST:
return HttpResponse('You requested a tip')
context = {}
return render(request, 'tip.html', context)
If the template containing the form is rendered by another view, you'll have to fix that view as well.
I am trying to make a button redirect the user to a new url after they submit a form. This is how it is right now, and it works properly and all the data gets sent to the django database.
<form method='POST' action='' class="col-md-6 col-md-offset-7" style="background-color: lightgreen; border-radius: 10px">
However, when I change the action to
action="{% url 'ridesharing:request-success' %}",
the redirect works, but the data does not go to my django database.
What is going on here?
You seem to have some confusion here. The action of the form is where the browser will send the data. Obviously, if you don't point that at the view which actually processes the data, then it won't be saved.
To redirect after a post, your view should return a redirect.
When you declare a form with an empty action attribute:
<form method='POST' action=''>
the POST data will be sent to the same URL. This is useful because if there is some error in the form, it's easy to re-display it with all fields filled with values entered by the user. Then, when the form becomes valid, a redirect is done to the confirmation page.
When you declare your form that way:
<form method='POST' action='{% url 'ridesharing:request-success' %}'>
The data entered by the user in the form will be sent to the view request-success. But this view probably only render the template of the confirmation page. If you want to correctly handle data from the form, you have to set action attribute of your <form> to the same view, or easier, keep it empty.
I do not understand why Daniel Roseman's post isnt accepted as the answer. It helped me when I wanted to redirect a create form to its update form.
Basically in the view of the app I defined the get_success_url to reverse_lazy to the data-update.
def get_success_url(self):
return reverse_lazy('app_name:data-update')
Just replace the 'app_name:data-update' with the appropriate url.
I am new to Django and I have a simple question. Here is a view :
def watchmovie(request, idmovie):
[...]
return render(request, 'movies/watch_movie.html', locals())`
and I would like to create a simple form :
an IntegerField that would redirect to the correct url :
if I submit "42" it will redirect me to the view watchmovie with the parameter 42 as idmovie.
How can I do that?
I tried something like that
<form action="{% url "movies.views.watchmovie" %}" method="post">
{% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Submit" />
</form>
my url.py is
from django.conf.urls import patterns, url
urlpatterns = patterns(
'movies.views',
url(r'^movie/(?P<idmovie>\d+)$', 'watchmovie'),
)
and Django says
Reverse for 'movies.views.watchmovie' with arguments '()' and keyword arguments '{}' not found. 1 pattern(s) tried: ['movies/movie/(?P<idmovie>\\d+)$']
Thank you!
The reason you are getting that error is because of a mistake in your url tag usage. Your watchmovie view url definition expects an argument to be supplied for idmovie. Since you are not supplying any argument in your url tag call, it looks only for urls which do not require an argument. Since there is none, you get an error.
But that is just a symptom. The real issue is that the way you have this structured there is no view listening for a post from your form.
The easier way to structure this is to use the same view to both display the form and to play the movie. If your view is hit with a GET request, display the form. If it is hit with a POST, validate the form (which will contain the movie id) and then respond with the page that plays the movie. That way there is no need to pass idmovie within your url.. you can remove that from your url definition and also remove the need to specify the action= attribute in your tag.. it will just post right back to where it came from.
I am using Django 1.5 and I want to check if the user has been redirected to my login page because he tried to access another page which required a login without being logged in. Suppose there is this view called 'securityPage' and suppose that view is called when I visit the
/securityInfo/
URL. Assuming this is the securityPage view:
#login_required
def securityPage(request):
#some code
Now when I visit that URL, it redirects me to the login.html page, which is correct (since the user - me - is not logged into any account). This is my login.html:
{% if 'next' in request.GET %}
<p>You must first login into your account before having access to the page you were trying to view.</p>
{% endif %}
for some reason, even though I am being redirected and the URL of the login page I am redirected to is
http://127.0.0.1:8000/login/?next=/securityInfo/
, the line
{% if 'next' in request.GET %}
evaluates to false even when 'next' is in the URL. Any idea why?
i don't think request object is part of the default context values.
check your template_context_processors in your settings and if you add django.core.context_processors.request in that list then the current request will be added to every requestcontext. or just pass the request manually for that view.
https://docs.djangoproject.com/en/1.5/ref/templates/api/#django-core-context-processors-request
When on a specific event page. The user can login using django social auth, but will redirect them to a static url, with the option below
SOCIAL_AUTH_LOGIN_REDIRECT_URL = '/dashboard/'
I was wondering if there is a dynamic way of redirecting to the page where they logged in. I tried this, but it didn't work.
{% url socialauth_begin 'facebook' %}?redirect_uri={{ request.get_full_path }}
Any suggestions would be great thanks.
Matthew
According to source code all you need is to set ?next=.
Like {% url socialauth_begin 'facebook' %}?next={{ request.get_full_path }} or another field name if setted REDIRECT_FIELD_NAME