Is there a way to render a html page without view model? - django

Is there a way to render a html page without having a view model in django if a page is going to display only static html?
Also, can I redirect to a html page instead of a url? For example, instead of doing this:
return HttpResponseRedirect('form/success/')
can I do this:
return HttpResponseRedirect('success.html')
?

direct_to_template no longer works in Django 1.8. Here is how to do it in 1.8, in the urls.py:
from django.views.generic import TemplateView
urlpatterns = [
url(r'^$', TemplateView.as_view(template_name="your_static.html"), name='whatever'),
]

You can render a template without a view using the generic view direct_to_template.
Redirecting to success.html isn't the way to go in django since success.html doesn't have an associated URL. You always have to bind the template to an url via a view (direct_to_template is a view which gets all its arguments from the conf in urls.py, but it's still a view)

For static HTML you could use Flatpages app.
You could also serve rendered templates (which could contain only static HTML) but you will need a view:
from django.shortcuts import render_to_response
def some_view(request):
return render_to_response('sometemplate.html')
About redirection, basically you can't redirect to HTML page by just giving the filename, even if it were statically served by the web server you would still be using a URL that points to that file.

I'm pretty sure this isn't what OP wanted to do, but if someone wants to render something without a view, (for sending an email or appending to other html code), you can use this:
from django.template.loader import render_to_string
html = render_to_string('template.html', params)

Use lambda function.
urlpatterns = [
path('', lambda request: render(request, 'homepage.html'))
]

Related

Two views same URL

I have a single dashboard_view URL path("", view=dashboard_view, name="dashboard").
On this page you can see the homepage unauthenticated. However, if you login, I present a modal popup to allow a user to populate a CreateForm.
The issue is that the dashboard_view doesn't have the form ( I have that in another view ). What is the best practice for this? Best for the user to have different options on the same page without having to switch pages.
You can use the login_required decorator. In login_required login_url is an optional parameter if you have declared the login path in the settings.py file. If your entire site has a login URL is the same. You can put LOGIN_URL = 'login_form_url' in the settings.py file.
from django.contrib.auth.decorators import login_required
#login_required(login_url='/login_form_url/')
def dashboard_view(request):
return render(request,'app_name/dashboard.html')

Why don't we pass any argument a function defined in 'views.py' - Django

In a Django project, when we define a function in viwes.py, it's expecting a one argument (something called request argument).
views.py
from django.http import HttpResponse
def my_homepage_view(request):
return HttpResponse("<h1>This is the homepage")
But in url.py, when we pass my_homepage_view in to url()function, we don't pass any argument to my_homepage_view() function. In this case, I didn't pass any argument to my_homepage_view(). But it worked fine.
url.py
from lern_django.views import my_homepage_view
urlpatterns = [
url(r'^$', my_homepage_view)
]
Please can you explain me how was that possible?
The url method you're using here is configuring the Django application to use that view function for that url. When Django processes a request, it will parse the HTTP request into a HttpRequest object and pass that to your view function.
In a simple word, when you request a page in your application, django itself creates
HttpRequest object and binds it to the responsible views as a first argument.
Go to the settings.py file and find following:
ROOT_URLCONF = 'project.urls'
When you request a page in django application following happens:
1) Django will look at your settings.py file and find ROOT_URLCONF and determines, which urls to follow
2) Every HTTPRequest from your browser will have URLCONF attribute associate with it.
3) Based on this URLCONF value in your HTTPRequest from browser, django map it to the ROOT_URLCONF
4) Django then look for urlpatterns in your urls and iterate over all the urls and exist on first occurence of given url pattern.
5) And, based on url mapping, given view will be loaded. The view will be called with HTTPRequest as a first parameter.

How to redirect to another page

Well, I have my own session and authentication routines in my app. When a user goes to app/login/ page, if he or she is not loggen in, then he/she sees a form to submit, otherwise I want to redirect the user to the root page app/. login view looks like this:
def index(request):
if sessionCheck(request):
# here I want to redirect one level above to app/
pass
else:
template = loader.get_template('login/login.html')
... other code which is not important
In PHP, I think, I would do it like this:
header("Location: /");
exit;
But how should I do exactly the same thing in Django?
You can use the redirect shortcut function:
from django.shortcuts import redirect
return redirect('some-view-name')
You can also specify a URL apart from a named view. So if your index url does not have a name, you can specify '/'.
This shortcut function is essentially the same as:
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse
return HttpResponseRedirect(reverse('some-view-name'))
In order to have a named view, in your urls.py you have to provide a name argument. For instance:
urlpatterns = [
url(r'^$', views.index, name='index'),
url(r'^some-view$', views.something, name='some-view-name'),
]
More on that on Django tutorial, Django URL dispatcher.
You can use redirect
if sessionCheck(request):
redirect(your_url)
else:
...

using normal html files instead of Django templates

I have a bunch of html files. I'd like to render them in my Django View without converting them to Django templates. Is this possible?
Or is there any code to do the conversion easily?
Actually, you don't have to write a view for such cases. You can use direct_to_template shortcut in urls.py as is.
From generic views doc
from django.conf.urls import patterns, url, include
from django.views.generic.simple import direct_to_template
urlpatterns = patterns('',
('^about/$', direct_to_template, {
'template': 'about.html'
}),
)
It is possible, just use the following method call:
def main_page(request):
template=get_template('index.html')
variables=Context({})
output = template.render(variables)
return HttpResponse(output)
Here your index.html file is located in templates folder.
Django templates are built on top of plain old files. If you have HTML files that don't have {{ or {% in them, then they are automatically valid Templates, albeit consisting of a single Text node.
Just reference them as you would reference any template, and it will "just work".
return render(request, "my-plain-html-file.html")

Django static page?

I want to make a static page which will be shown to the user only if he/she clicks on a link provided in one of my models. I can do this by making a Python page alone and calling it, but I want it be called from Django. The user interface should be constructed using the Django API only.
Any suggestions?
With the class-based views in newer Django versions, one can use this in urls.py:
from django.views.generic import TemplateView
url(r'^about',
TemplateView.as_view(template_name='path/to/about_us.html'),
name='about'),
Bypassing views to render a static template, add this line in "urls.py". For example "About Us" page could be
(r'^about', 'django.views.generic.simple.direct_to_template', {'template': 'path/to/about_us.html'}),
Do you mean something like Django's flatpages app? It does exactly what you describe.
If you want to make a static page the flatpages is a good choice. It allows you to easily create static content. Creating static content is not harder than creating a view really.
On Django 2.2.6, loosely following David's answer, I added the path in urls.py:
from django.views.generic import TemplateView
urlpatterns = [
.... .... ....
path('about',
TemplateView.as_view(template_name='path/to/about_us.html'),
name='about'),
And I needed to adjust settings.py to specify the template directory:
TEMPLATES = [{
... ... ...
'DIRS': [os.path.join(BASE_DIR, 'template')],
Then I saved the actual content in template/path/to/about_us.html