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Can Django's URL tag show the URL of the RedirectView

Let's say that in my urls.py I have a url like this:
path("support/", RedirectView.as_view(url="http://www.example.com"), name="support"),
And in one of my templates I use the url tag:
{% url "support" %}
This of course outputs /support/ as expected. But what if I want it to output http://www.example.com instead? Is that at all possible? Skip the redirect basically.
So Link would output Link.

But what if I want it to render http://www.example.com instead? Is that at all possible? Skip the redirect basically.
No, in short it is not possible with django views, since the url is of another website and you can't render it in your own.
You can see what exactly render() does.
If you'd like to directly redirect, then simply use anchor tag as:
Visit website.
You can also do this dynamically, by creating a URLField in one of the models and simply iterating it with href attribute of anchor tag.

Cannot have any URLs with slugs. NoReverseMatch

I'm a begginer grasping at straws with difficulty dealing with the django slug url system and these NoReverseMatch errors that make no sense to me even after reading the docs.
I have a django project. In one of the views, I pass a list of geoJSON features into a template, and show them on a map. I want to have each feature act as a clickable 'link' to a view that will show stuff about it. The following is part of the template that has those features that I want to click on:
//part of the template:
<script type="text/javascript">
...
function onEachFeature(feature, layer) {
layer.on('click', function (e) {
window.location.href = "{% url 'polls:areadetail' feature.properties.myslug%}";
});
}
(I have confirmed that feature.properties.myslug does in fact contain the slug I want).
The url pattern I want to go to:
urlpatterns = [...
url(r'^areadetail/(?P<areaslug>[-\w]+)/$', views.AreaDetail, name='areadetail'),]
And the view it relates to:
def AreaDetail(request, areaslug):
area = get_object_or_404(Area, nameslug=areaslug)
return render(request, 'polls/areadetail.html', {'area':area})
The issue I get is, by doing what I show and placing that url reference inside that template I show above, that I want to be able click on, that template won't even work at all, giving me a 'Error during template rendering' full page error info that starts with:
NoReverseMatch at /polls/areas/
Reverse for 'areadetail' with arguments '('',)' and keyword arguments '{}' not found. 1 pattern(s) tried: [u'polls/areadetail/(?P[-\w]+)/$']
Any help would be immensely appreciated
EDIT part1: As I've said in response to falsetru, I'm sure feature.properties.myslug has in fact got a slug expression in it.
EDIT2: Based on something I found in a django ticket, I've made a slight change in the url regex at urls.py, from (?P<areaslug>[-\w]+)/$ to (?P<areaslug>[-\w]+)?/$ and now the error is:
Page not found (404)
Request Method: GET Request URL: http://127.0.0.1:8000/polls/areadetail// Raised by: polls.views.AreaDetail
Is it possible that because the "{% url 'polls:areadetail' feature.properties.myslug%}" bit is inside javascript, that feature.properties.myslug is not being inserted there correctly? Like some sort of brackets are needed here?

According to the error message, feature.properties.myslug is empty or has no value.
Make sure the feature.properties.myslug is passed correctly from view.
Comment out {% url .. %} temporarily.
Print {{ feature }}, {{ feature.properties }}, {{ feature.properties.myslug }} to see if which part is missing.
Fix view accordingly.
Uncomment {% url .. %}.

After some more digging around I've found the answer to why doesn't this work in another question at:
How to pass javascript variable to django custom filter
The answer to it by Ludwik Trammer says:
Django templates are build on the server side, while JavaScript is executed on the client side.
That means that template code is always executed before JavaScript (as
it is executed by the server, before the page is sent to the client).
As a consequence it is absolutely impossible to mix JavaScript and
Django code the way you want to.
Which clearly applies here. I was focused on problems with the URL template, regex on the urls.py file etc. when the problem was that no matter what I did, because it's in a javascript section, run client-side, that URL template will always be incomplete no matter what I do, therefore being an impossible solution to what I want.

Managing django urls

I am making a personal site. I have a blog page(site.com/blog), where I have a list of my blog posts. If I want to check a blog post simple enough I can click it and the code:
<h1>{{ obj.topic_title }}</h1>
will get me there. Also if I want to go to my contacts page (site.com/contacts). Easy enough I click nav contacs button and I go there
Contacts
but if I enter a blog post (site.com/blog/1), I am using the same template and if I want to go to my contacts page I have to yet again click the
Contacts
link, but that will port me to a 404 page site.com/blog/contacts . How do I deal with this problem without harcoding every single page

Use the built-in Django url template tag, which takes the view name and returns an absolute path to it.
Returns an absolute path reference (a URL without the domain name) matching a given view and optional parameters.
You can give it a view name and any of its view parameters as well. This is much better than how you link to the blog page:
{% url 'blog-view-name' obj.id %}
This ensures that if you ever change the structure of your views, it will still not break your links.

Using variables in Django model content

i like to use the url template tag in my model's content.
example:
models content:
Car.description = 'this is a link to our main page: home'
in template.html:
<div>{{ Car.description }}</div>
result
<div>this is a link to our main page: home
is it possible, or do i have to write my own template tag?
thanks in advance
Roman

Assuming you have this:
car.description = 'this is a link to our main page: home'
You can do:
from django.template import Context, Template
from django.core.urlresolvers import reverse
class Car(models.Model):
def description_with_url(self):
return Template(self.description).render({'url': reverse('home')})
or use the same logic in custom template tag instead of method..

I can't figure out why you would need to do that. Assuming that I fully-understood your question, you are attempting to store something within a model's field that then behaves "dynamically" when rendered.
A model field's content that stores a URL should contain a URL and only a URL by utilizing the URLField.
Else, if you're dynamically building the URL from somewhere else, simply use template markup, i.e. the url template tag as it is meant to be used; it can also take parameters depending on the specific URL pattern. I.e. the url tag is meant to be used in the template's context.
Let me know if you update the question to describe what you are trying to achieve. But storing "behaviour" at data level is something to simply stay away from.

How to get the current URL of a page in a Django template?

I know there is another question with virtually the same title as mine but the solution in that one didn't work for me. My url is like this:
http://domain.com/videos/dvd/1/
If I use either {{baseurl}} or {{ request.get_full_path }} I get just this part:
http://domain.com/videos/
How can I get the entire url? I need to be able to do this from the template level.
EDIT
P.S. it should disregard any parameters that may be in the url.

You could get it in your view and pass it along into your template context so that it is available to you there.
https://docs.djangoproject.com/en/1.3/ref/request-response/#django.http.HttpRequest.build_absolute_uri
full_url = request.build_absolute_uri(None)
# pass full_url into the template context.