In my Django project, I have created a custom admin page for an app via the get_urls() method. I'd like to add a link to the app's main model index view that will take users to this custom page - however, I'm having some trouble creating this link element correctly and I don't seem to be able to piece together the right way to do it - I'm just left with a Reverse for 'export' not found. 'export' is not a valid view function or pattern name. error.
I've set up the admin for the app like so:
# my_project/observations/admin.py
from django.template.response import TemplateResponse
from django.urls import path
class ObservationAdmin(SimpleHistoryAdmin, SoftDeletionModelAdmin):
change_list_template = 'export_link.html'
def get_urls(self):
urls = super().get_urls()
custom_urls = [
path('export/', self.admin_site.admin_view(self.export_view), name='export')
]
return custom_urls + urls
def export_view(self, request):
context = dict(
self.admin_site.each_context(request),
)
return TemplateResponse(request, 'export.html', context)
and the two templates that are referenced:
# my_project/observations/templates/export.html
{% extends "admin/base_site.html" %}
{% block content %}
<div>
Some custom content
</div>
{% endblock %}
# my_project/observations/templates/export_link.html
{% extends 'admin/change_list.html' %}
{% block object-tools-items %}
<li>
Export
</li>
{{ block.super }}
{% endblock %}
Navigating directly to http://localhost:8000/admin/observations/observation/export/ works perfectly, I see the custom content page exactly as I want it... so the issue I'm striking is with the link template - I get the Reverse... error when I navigate to the model index page.
Perhaps the argument I'm passing to url is incorrect, or I need to register that URL elsewhere - but I don't quite know. The other examples of link elements like this that I've been able to find don't reference URLs created via the admin class' get_urls() method - so any guidance on this would be greatly appreciated.
Thanks very much, let me know if there's any other info that I can provide to help sort this out.
I think the problems is in missing namespace in your export_link.html template. Instead of:
Export
try:
Export
I'm trying to send an object obtained with a ModelChoiceField into a view via urls and views parameters by I don't think that's the right method. I don't know if I have to use POST or GET method.
When I use the GET method, the object is displayed in the url but the view stay the same.
With the POST method, django send errors messages about parameters.
EDIT : I don't remeber the exact messages, I can't recover them for now but as I remember they said something like Reverse for argument not found
My form
class IndividuForm(forms.Form):
individu = forms.ModelChoiceField(queryset=Individu.objects.all()
Here's the view with the form
def index(request):
individu_form = IndividuForm(request.GET or None)
if individu_form.is_valid():
individu_id = individu_form.cleaned_data['individu'].id
HttpResponseRedirect('%i/raison/' % individu_id)
return render(request, 'derog_bv/index.html', {'individu_form':individu_form})
The template where the form is displayed
<form action="{% url 'index' <!-- Errors when I put parameters here --> %} method="<!-- GET or POST -->">
{% csrf_token %}
{{ form }}
<input type="submit">Suivant</input>
</form>
The view where I want to get my object
def raison(request, """ error with this parameter : individu_id"""):
individu = get_object_or_404(Individu, pk=individu_id)
URLs
urlpatterns = [
path('', views.index, name='index'),
path('<int:individu_id>/raison/', views.raison, name='raison'),
]
Ok, so:
1/ you definitly want to use the GET method for your form (you're not submitting data for processing / server state change)
2/ I don't know why you'd want to pass parameters (nor which parameters FWIW) to the {% url %} tag in your index template - you're submitting the form to the index view, which doesn't expect any additional param. Actually you could just remove the action attribute of the HTML form tag since the default (submitting to the current url) is what you want.
3/ your views.raison prototype must match the url definition, so it has to be:
def raison(request, individu_id):
# code here
4/ in your index view:
HttpResponseRedirect('%i/raison/' % individu_id)
you want to build the url using django.core.urlresolvers.reverse instead :
from django.core.urlresolvers import reverse
def index(request):
# code here
if ...:
next = reverse("raison", kwargs={"individu_id": individu_id})
return HttpResponseRedirect(next)
or - even easier - just use the redirect shortcut:
from django.shortcuts import redirect
def index(request):
# code here
if ...:
return redirect("raison", individu_id=individu_id)
There might be other issues with your code but from the infos you posted, those hints should at least put you back on tracks.
Django newbie here. Following the documentation, I am trying the following to get a link to the admin site from the homepage of the public site I'm building:
{% load admin_urls %}
<p>Go to the admin.</p>
I am getting the error:
NoReverseMatch at /
u'opts|admin_urlname' is not a registered namespace
I am including the URLs properly:
url(r'^admin/', include(admin.site.urls)),
My template loaders are in the right order.
I've tried a few different variations on this, and they all throw namespace errors.
Any ideas? Thanks!
After 30 minutes with Daniel Roseman / Django docs in one screen and my code in the other, I come up with this simple solution:
In your views.py, add the opts context with the _meta of the model (that includes the required app_label and model_name):
class YourModelDetailView(DetailView):
def get_context_data(self, **kwargs):
context = super(YourModelDetailView, self).get_context_data(**kwargs)
context["opts"] = YourModel._meta
return context
In your templates:
{% url opts|admin_urlname:'change' object.pk %}
Where change can be any action in the reverse admin urls documentation page.
While the above answers were helpful about the code I was calling, there is a much easier way. I'm using this instead:
{% url 'admin:index' %}
This works for custom admin views as well, like:
{% url 'admin:myapp_mymodel_<keyword>' object.id %}
Where keyword is from the named parameters listed here (i.e. add, change, delete).
You are almost certainly using the released 1.4 version, rather than the development version. As the documentation for that version shows, you need to use {% load url from future %} before you can use that syntax.
I have a url defined as follows:
url(r'^details/(?P<id>\d+)$', DetailView.as_view(), name='detail_view'),
In my templates, I want to be able to get the following url: /details/ from the defined url.
I tried {% url detail_view %}, but I get an error since I am not specifying the id parameter.
I need the url without the ID because I will be appending it using JS.
How can I accomplish this?
Just add this line to your urls.py:
url(r'^details/$', DetailView.as_view(), name='detail_view'),
or:
url(r'^details/(?P<id>\d*)$', DetailView.as_view(), name='detail_view'),
(This is a cleaner solution - thanks to Thomas Orozco)
You'll need to specify that id is optional in your view function:
def view(request, id=None):
How can I get the full/absolute URL (e.g. https://example.com/some/path) in Django without the Sites module? That's just silly... I shouldn't need to query my DB to snag the URL!
I want to use it with reverse().
Use handy request.build_absolute_uri() method on request, pass it the relative url and it'll give you full one.
By default, the absolute URL for request.get_full_path() is returned, but you can pass it a relative URL as the first argument to convert it to an absolute URL.
>>> request.build_absolute_uri()
'https://example.com/music/bands/the_beatles/?print=true'
>>> request.build_absolute_uri('/bands/?print=true')
'https://example.com/bands/?print=true'
If you want to use it with reverse() you can do this : request.build_absolute_uri(reverse('view_name', args=(obj.pk, )))
If you can't get access to request then you can't use get_current_site(request) as recommended in some solutions here. You can use a combination of the native Sites framework and get_absolute_url instead. Set up at least one Site in the admin, make sure your model has a get_absolute_url() method, then:
>>> from django.contrib.sites.models import Site
>>> domain = Site.objects.get_current().domain
>>> obj = MyModel.objects.get(id=3)
>>> path = obj.get_absolute_url()
>>> url = 'http://{domain}{path}'.format(domain=domain, path=path)
>>> print(url)
'http://example.com/mymodel/objects/3/'
https://docs.djangoproject.com/en/dev/ref/contrib/sites/#getting-the-current-domain-for-full-urls
You can also use get_current_site as part of the sites app (from django.contrib.sites.models import get_current_site). It takes a request object, and defaults to the site object you have configured with SITE_ID in settings.py if request is None. Read more in documentation for using the sites framework
e.g.
from django.contrib.sites.shortcuts import get_current_site
request = None
full_url = ''.join(['http://', get_current_site(request).domain, obj.get_absolute_url()])
It isn't as compact/neat as request.build_absolute_url(), but it is usable when request objects are unavailable, and you have a default site url.
If you don't want to hit the database, you could do it with a setting. Then, use a context processor to add it to every template:
# settings.py (Django < 1.9)
...
BASE_URL = 'http://example.com'
TEMPLATE_CONTEXT_PROCESSORS = (
...
'myapp.context_processors.extra_context',
)
# settings.py (Django >= 1.9)
TEMPLATES = [
{
'BACKEND': 'django.template.backends.django.DjangoTemplates',
'DIRS': [],
'APP_DIRS': True,
'OPTIONS': {
'context_processors': [
'django.template.context_processors.debug',
'django.template.context_processors.request',
'django.contrib.auth.context_processors.auth',
'django.contrib.messages.context_processors.messages',
# Additional
'myapp.context_processors.extra_context',
],
},
},
]
# myapp/context_processors.py
from django.conf import settings
def extra_context(request):
return {'base_url': settings.BASE_URL}
# my_template.html
<p>Base url is {{ base_url }}.</p>
In your view, just do this:
base_url = "{0}://{1}{2}".format(request.scheme, request.get_host(), request.path)
This worked for me in my template:
{{ request.scheme }}://{{ request.META.HTTP_HOST }}{% url 'equipos:marca_filter' %}
I needed the full url to pass it to a js fetch function.
I hope this help you.
django-fullurl
If you're trying to do this in a Django template, I've released a tiny PyPI package django-fullurl to let you replace url and static template tags with fullurl and fullstatic, like this:
{% load fullurl %}
Absolute URL is: {% fullurl "foo:bar" %}
Another absolute URL is: {% fullstatic "kitten.jpg" %}
These badges should hopefully stay up-to-date automatically:
In a view, you can of course use request.build_absolute_uri instead.
Yet another way. You could use build_absolute_uri() in your view.py and pass it to the template.
view.py
def index(request):
baseurl = request.build_absolute_uri()
return render_to_response('your-template.html', { 'baseurl': baseurl })
your-template.html
{{ baseurl }}
Examine Request.META dictionary that comes in. I think it has server name and server port.
To create a complete link to another page from a template, you can use this:
{{ request.META.HTTP_HOST }}{% url 'views.my_view' my_arg %}
request.META.HTTP_HOST gives the host name, and url gives the relative name. The template engine then concatenates them into a complete url.
Try the following code:
{{ request.scheme }}://{{ request.META.HTTP_HOST }}
If you're using django REST framework, you can use the reverse function from rest_framework.reverse. This has the same behavior as django.core.urlresolvers.reverse, except that it uses a request parameter to build a full URL.
from rest_framework.reverse import reverse
# returns the full url
url = reverse('view_name', args=(obj.pk,), request=request)
# returns only the relative url
url = reverse('view_name', args=(obj.pk,))
Edited to mention availability only in REST framework
I know this is an old question. But I think people still run into this a lot.
There are a couple of libraries out there that supplement the default Django functionality. I have tried a few. I like the following library when reverse referencing absolute urls:
https://github.com/fusionbox/django-absoluteuri
Another one I like because you can easily put together a domain, protocol and path is:
https://github.com/RRMoelker/django-full-url
This library allows you to simply write what you want in your template, e.g.:
{{url_parts.domain}}
If anyone is interested in fetching the absolute reverse url with parameters in a template , the cleanest way is to create your own absolute version of the {% url %} template tag by extending and using existing default code.
Here is my code:
from django import template
from django.template.defaulttags import URLNode, url
register = template.Library()
class AbsURLNode(URLNode):
def __init__(self, view_name, args, kwargs, asvar):
super().__init__(view_name, args, kwargs, asvar)
def render(self, context):
url = super().render(context)
request = context['request']
return request.build_absolute_uri(url)
#register.tag
def abs_url(parser, token):
urlNode = url(parser, token)
return AbsURLNode( urlNode.view_name, urlNode.args, urlNode.kwargs, urlNode.asvar )
Usage in templates:
{% load wherever_your_stored_this_tag_file %}
{% abs_url 'view_name' parameter %}
will render(example):
http://example.com/view_name/parameter/
instead of
/view_name/parameter/
I got it:
wsgiref.util.request_uri(request.META)
Get the full uri with schema, host, port path and query.
You can either pass request reverse('view-name', request=request) or enclose reverse() with build_absolute_uri request.build_absolute_uri(reverse('view-name'))
Not for absolute url but I was looking just to get host. If you want to get host in your view.py you can do
def my_view(request):
host = f"{ request.scheme }://{ request.META.get('HTTP_HOST') }"
As mentioned in other answers, request.build_absolute_uri() is perfect if you have access to request, and sites framework is great as long as different URLs point to different databases.
However, my use case was slightly different. My staging server and the production server access the same database, but get_current_site both returned the first site in the database. To resolve this, you have to use some kind of environment variable. You can either use 1) an environment variable (something like os.environ.get('SITE_URL', 'localhost:8000')) or 2) different SITE_IDs for different servers AND different settings.py.
Hopefully someone will find this useful!
While working on a project I came to know to get the full/absolute URL in Django.
If your URL looks like this in the address bar:
https://stackoverflow.com/questions/2345708
And if you want to show the above URL to your template.
{{ request.path }} #Without GET parameters.
{{ request.get_full_path }} #with GET parameters
For the above two codes, this will print in your template will be
questions/2345708
and another way to get a full URL is:
{{request.build_absolute_uri}}
this will print in your template will be:
https://stackoverflow.com/questions/2345708
There is also ABSOLUTE_URL_OVERRIDES available as a setting
https://docs.djangoproject.com/en/2.1/ref/settings/#absolute-url-overrides
But that overrides get_absolute_url(), which may not be desirable.
Instead of installing sites framework just for this or doing some of the other stuff mentioned here that relies on request object, I think the better solution is to place this in models.py
Define BASE_URL in settings.py, then import it into models.py and make an abstract class (or add it to one you're already using) that defines get_truly_absolute_url(). It could be as simple as:
def get_truly_absolute_url(self):
return BASE_URL + self.get_absolute_url()
Subclass it and now you can use it everywhere.
I came across this thread because I was looking to build an absolute URI for a success page. request.build_absolute_uri() gave me a URI for my current view but to get the URI for my success view I used the following....
request.build_absolute_uri(reverse('success_view_name'))
<div class='col-12 col-md-6'>
<p class='lead'>Login</p>
{% include 'accounts/snippets/form.html' with form=login_form next_url=request.build_absolute_uri %}
</div>
Here for example am saying load the form and tell the form that the next URL is the current URL which this code rendred from
I use this code :
request.build_absolute_uri('/')[:-1]
response :
https://yourdomain.com
request.get_host() will give you the domain.
request.get_host()
Use this for request object for APIView in django
class WalletViewSet(mixins.ListModelMixin, GenericViewSet):
serializer_class = WalletSerializers
pagination_class = CustomPaginationInvestment
def get_queryset(self):
######################################################
print(self.request.build_absolute_uri())
#####################################################
wallet, created = Wallet.objects.get_or_create(owner=self.request.user)
return Wallet.objects.filter(id=wallet.id)
You get output like this
http://localhost:8000/v1/wallet
HTTP GET /v1/wallet 200 [0.03, 127.0.0.1:41608]
You can also use:
import socket
socket.gethostname()
This is working fine for me,
I'm not entirely sure how it works. I believe this is a bit more low level and will return your server hostname, which might be different than the hostname used by your user to get to your page.
You can try "request.get_full_path()"