I have a lot of apps running on my site and I was wanting to make all the views accept a certain kwarg without having to go in and edit them all manually? Is there a way to do this?
I suppose I should add it into the django base view class somewhere, but I am unsure exactly where to add it to in that?
Any ideas?
EDIT:
What I am trying to do is have translations set in my db under a certain model and then have the site default text areas be displayed in a certain language based on the url...
/eng/some/url
/esp/some/url
those two urls would display different languages, but I have to capture the variable and put it into each and every view which is quite cumbersome
Django already has some i18n support in urls, check it out. You need to activate django.middleware.locale.LocaleMiddleware by adding it to your settings.MIDDLEWARE_CLASSES and to tune your urlconf a bit by wrapping your urls with i18n_patterns.
The complete example is given in the docs, I see no sense copying it here.
Related
Lets say I have installed a 3rd party app called 'articles', the app contains basic templates and views. And there is a view called 'home' to list articles.
I need to add a form within that view and of course the form variable is not in the default 'home' view. How should I go about adding the form variable to that view?
There are a couple of ways I can think of right now:
Create another app and create a custom view.
This seems crazy and I won't do it, but for the sake of possibility, add a context processor to add the form variable into the context.
Just wondering if anyone had this situation and what is the best approach for this?
well, you may want to try a Function Decorator to redefine the previous view function.
the context processor or the middlerware is not recommend because it's global and dirty.
another use way I can thing is use-defined tags. This may takes more affort and go against the origin design of the tags. But it seem to be a good way.
I want to implement a report section in Django admin. This would mean adding a custom section in the admin homepage where instead of a list of models I would see a list of reports. I want to use Django's admin tables with filters, sorting, everything if possible.
What would be the "best" way of achieving this? I realize this is a "big" question so I'm not asking for code snippets necessarily, a summary of needed actions would be just fine :)
P.S. Be report I mean a "made up" model by custom queries (queryset or how it's called).
P.S.2 Maybe this question should be something like: How to use Django admin tables functionality in own admin view?
P.S.3 Or maybe there is a way of providing to the existing admin interface my own data. This way I don't have to do anything else. I just want to say instead of a model take this data and display it in a nice table which I can sort, filter etc etc.
So you are attempting to add in new pages into the django admin.
This section explains to you exactly how you can do so - https://docs.djangoproject.com/en/dev/ref/contrib/admin/#adding-views-to-admin-sites
The basic idea is to add in new urls that you want in your urls.py as if you are adding urls for your "front end" pages. The key difference is that these new urls you are adding should start with ^admin/ and would look something like ^admin/my_special_link_in_admin and this url will point to your own custom view function at a location you so prefer.
E.g.
(r'^admin/my_special_link_in_admin/$', 'my_custom_admin_app.views.special_admin_page'),
So this is the way for complete customization. There's a very good tutorial which I refer to here - http://brandonkonkle.com/blog/2010/oct/4/django-admin-customization-examples/
In addition, if you don't want to do too much work, consider using Django Admin Plus - https://github.com/jsocol/django-adminplus
Or a django-admin-views - https://github.com/frankwiles/django-admin-views
My situation is as follows.
I have a django app that is a CMS, and this one app generates page content and menus.
There is 'default' view that generates all the page content. One of the fields of the main model is designed to limit the visibility of pages depending on the setting:
'internal' (limited to users on our network),
'worldwide' (viewable by the www) and now, I would like to add an extra value
'secure', which would limit viewing only to those who are logged in to the site.
I dont think #login_required would work, as it only works on whole functions.
We want the author or editor to be able to set this flag, rather than code a special function; and there will be instances where a page migrates from 'internal' to 'secure' then to 'worldwide' - so ideally the urls should stay the same.
What is the best way to go about this?
thanks in advance... patrick
As you want the flag set on your object, I'm assuming that flag can't be read until you're already within the view (i.e. you won't be storing it in something accessible by the request object, like a session), so that precludes custom decorators. You could choose to go with something a bit crude but workable like this:
if (val=="internal" and inrange(request.META['REMOTE_ADDR'])) or (val=="secure" and request.user.is_authenticated()) or val=="worldwide":
return render_to_response .... etc.
else:
return HttpResponseForbidden()
Substituting your model values, and writing the inrange function, naturally.
The more sophisticated approach, though, would be to write custom middleware and use the process_view() function. See the docs for more details.
I have just learnt about Django apps. I want to know that within one site, if I make different apps. like, users, profiles, polls, blogs,comments, jobs , applications then how can I manage them to make them intereactive? And is it how the concept of app should be? I want to have things loosely coupled that's why asking? Rails work on the way of REST, so do Django support that also with the use of apps? May be my questions seems a bit ambiguous because I am new to django and some of my concepts are still messed up.
Please tell what ever you know.
The general idea is that apps should be as loosely coupled as possible. The goal is to have completely self-contained functionality. Now, of course, that's not always possible and many times it even makes sense to bring in functionality from another app. To do that, you simply import whatever you need. For example, if your "blogs" app needed to work with your Comment model in your "comments" app you'd simply add the following to the top of the python file you're working in:
from comments.models import Comment
You can then use Comment as if it were defined right in the same file.
As far as REST goes, Django's views are much more fluid. You can name your view whatever you like; you need only hook it up to the right urlpattern in urls.py. Django views can return any content type, you just prepare the response and tell it what mimetype to serve it as (the default is HTML).
I'm working on a Django site with a basic three column design. Left column navigation, center column content and right column URL specific content blocks.
My question is about the best method of controlling the URL specific content blocks in the right column.
I am thinking of something along the lines of the Flatpages app that will make the content available to the template context if the URL matches a pre-determined pattern (perhaps regex?).
Does anyone know if such an app already exists?
If not, I am looking for some advice about the best way to implement it. Particularly in relation to the matching of patterns to the current URL. Is there any good way to re-use parts of the Django URL dispatcher for this use?
Django CMS is a good suggestion, it depends on how deep you want to go. If this is just the beginning of different sorts of dynamic content you want then you should go that way for sure.
A simple one-off solution would be something like this:
You would just need to write a view and add some variables on the end of the URL that would define what showed up there. Depending on how fancy you need to get, you could just create a simple models, and just map the view to the model key
www.example.com/content/sidecontent/jokes/
so if "jokes" was your block of variable sidecontent (one of many in your sides model instances) the urls.py entry for that would be
(r'^content/sidecontent/(?P<side>)/$,sides.views.showsides),
and then in your sides app you have a view with a
def showsides(request, side):
Sides.objects.get(pk=side)
etc...
For something like this I personally would use Django CMS. It's like flatpages on steroids.
Django CMS has a concept of Pages, Templates, and Plugins. Each page has an associated template. Templates have placeholders where you can insert different plugins. Plugins are like mini-applications that can have dynamic model-based content.
Although Django-CMS is an interesting suggestion, there are quite a few projects that do specifically what you've requested - render blocks of content based on a URL. The main one that I know about is django-flatblocks.