I'd like to know why is happening this.
I open my Django project (at 127.0.0.1:8000) and it appears my index
page as I've set it in URLs and views.
At the top I have a piece of code that if I'm authenticated it
appears some text, and if not, another one.
It appears when I open the index page that I'm not authenticated, but
when I click a link in my menu to other page in my project, it
appears that I'm actually authenticated.
I'm looking in the web for possible explanations but can't find out an answer to why I'm authenticated or logged in in a page and then in another not.
How could this be possible, and how could I manage it?
Check for #login_required decorator over your view functions. The documentation is at:
https://docs.djangoproject.com/en/1.9/topics/auth/default/#the-login-required-decorator
It is difficult to answer your question without seeing your code, but it sounds like you are not passing the request context to your index view. Make sure you are using the render shortcut in your views:
from django.shortcuts import render
def index(request):
return render(request, 'index.html')
Related
this is probably a noobie question.
I'm working on a Django project, after a request, I reload the page using:
return redirect('home')
Is there anyway to reload the current page without writing the url path? I don't need to specify it, since the page I need to reload is the one that is actually open.
Hope someone can help and sorry for easy english
You can use the following code :
from django.http import HttpResponseRedirect
return HttpResponseRedirect(request.path_info)
Your request contains the actual path so you can redirect to the same page without specifying explicitly
Path Info Django Doc
I am making a project where each client have their own app. This is because they have similar pages but not exactly the same so I think it's a good approach(I may be wrong) to just copy one app for each new client. I have not tried it yet, I am still planning for it. I see one problem with the view site link in the admin. I will let the clients use the admin. How can I set the view site link to the main page for the client? One way to solve it would be to leave it as is and have a function checking their user name and redirecting to the right app. But is there any other way to solve this problem?
I don't think it will be a good idea to have applications generated for users as that's too much once you reached a specific amount of users.
they have similar pages but not exactly the same
Then what you should do is to, after getting the user in your view, pass in a different context to your template. Something like:
def my_view(request):
# First assign different context to different users
context = {'data': 'whatever each user gets', 'other': 'put in more than 1 data',}
return render(request, 'myapp/index.html', context)
I will let the clients use the admin
That's not a good idea as the clients MUST be a superuser to view the admin site. Otherwise, you need to change the permissions of a superuser, make a separate superuser if you are the maintainer of the site, and all sorts of trouble. Just take some time and make your own templates.
I have been testing a few django voting apps and found qhonuskan-votes. I have managed to install it and is works great. However, I also want it allow voting rights to non authenticated users which I am not able to do. Need help with this please.
Here is the link for its models.py, views.py and compact.py files of this app.
models
views
compact
You could write a custom view which would look like def vote(request, model, object_id, value) from the external app, but without this piece of code in it:
if not request.user.is_authenticated():
return HttpResponse(status=401)
Also make sure, that you map your custom view to the correct url instead of including app's urls:
url(r'^vote/$', view='custom_vote', name='qhonuskan_vote'))
This is not the best solution, because you are simply rewriting the code from the external app and I can't think of any proper way to override the default view in a way that would suit your needs. A better solution would be to use a different app, which allows votes by unauthenticated users (if a few lines of additional code are not a problem you can use this).
I'm currently using out-of-the-box django.contrib.auth to handle authentication in my Django app. This means that the user starts at a log in page and is redirected to the app on successful login. I would like to make my app single-page, including this login process, where a redirect doesn't happen, but maybe a "hot" template switch-out or some fancy client-side div magic (that still remains secure). My Google searching turned up pretty short, the closest solution dealing with putting a log in form on every page.
Any direction or ideas here would be much appreciated. I would obviously prefer to work within the existing confines of django.contrib.auth if possible, but I'm open to all solutions.
I'm not sure I understand your question completely. I think you want to have a single page. If so, put logic in your template that checks to see if the user is authenticated. If not, display a login form that POSTS to the appropriate django.contrib.auth view. You can supply an argument to this view to have it redirect back to your page. When you come back, the user will be authenticated, so you won't display the login form.
Have a look at Django-Easy-Pjax https://pypi.python.org/pypi/django-easy-pjax - it works like a charm and is well documented. Everything you like is being made with AJAX requests: links, forms using GET and forms using POST.
Essentially you only need to add a data-pjax="#id_of_the_container_where_the_result_goes" attribute in your a and form tags.
And the great thing about it: It updates the title and location bar of your browser.
One caveat: If you want to upload files in some form, this is not supported by Easy-Pjax, so you might want to use some workaround jQuery library for that.
I am facing a problem: I want to give a link in my change form that will redirect to a page which may be simple php page also or any page, in that page I want to perform some db queries and display them. I also wan to pass id on click. Is it possible?
In my view.py I wrote:
from django.shortcuts import render_to_response
from django.template import RequestContext
def MyClass(self,id,request):
return render_to_response('admin/custom_change_form.html')#my template location
My model and admin files are simple.
To send to a file directly, use direct_to_template(). You can pass anything in the url that you like - just give the information to your template, and write it in the url. After all, Django doesn't require url helpers.
I sense that whatever you're trying to do is some god-awful hackish thing that would be much better served by doing it all in Django.
You will need to override your change form for that model and display whatever you would like. But that is making the assumption you are talking about the contrib admin within Django.
Realistically you have not provided sufficient information for anyone to accurately answer your question.