I am trying a tutorial on Django called blog. I have the following structure:
FirstBlog|FirstBlog
settings
urls
__init__
etc
blog
templates | index.html
migrations
views.py
manage.py
The view.py has
from django.shortcuts import render
from django.shortcuts import render_to_response
from blog.models import posts
def home(request):
return render('index.html')
The urls.py has
from django.conf.urls import url
from django.conf.urls import include
from django.contrib import admin
urlpatterns = [
url(r'^blog', 'FirstBlog.blog.views.home',name='home'),
]
and I get this error:
Using the URLconf defined in FirstBlog.urls, Django tried these URL patterns, in this order: ^blog [name='home']
The current URL, , didn't match any of these.
I can't seem to get it right..
Any help would be appreciated.
Thank you,
You are requesting for / url and you have not saved any such mapping. Current mapping is for /blog . So it will work for the same url.
i.e goto the browser and request /blog
If you need it to work for / then change the urls appropriately.
within your blog app, create a urls.py file and add the following code which calls the home view.
blog/urls.py
from django.conf.urls import url
from . import views
urlpatterns = [
#url(r'^',views.home, name='home'),
]
then in your root urls file which can be found at FirstBlog/urls.py
from django.conf.urls import url, include
from django.contrib import admin
urlpatterns = [
url(r'^blog/',include('blog.urls')), #when you visit this url, it will go into the blog application and visit the internal urls in the app
]
PS:
your templates should be in blog/templates/blog/index.html
Read this docs on templates to understand how django locates templates.
This one is to understand how urls work Urls dispatcher
You are doing this in the wrong way! Try doing that using TemplateView from class-based views, which are the current standard from django views.
Check the django documentation: https://docs.djangoproject.com/en/1.9/topics/class-based-views/
Use this at the urls.py:
from django.conf.urls import url
from django.views.generic import TemplateView
urlpatterns = [
url(r'^blog/', TemplateView.as_view(template_name="index.html")),
]
And then, create a folder called templates in the project root ( like this: https://docs.djangoproject.com/en/1.9/intro/tutorial03/#write-views-that-actually-do-something ) named index.html
Simply go to file then select Save all your project instead of save. Or use shortcut Ctrl +k s on windows. Project should be able to sync and display the code on Django interface
Related
I currently have a website where the home page at www.mysite.com is the wagtail blog index page
I wish to move the blogindex page to another url
I can easily have a different homepage by amending my urls.py file:
#original
path("", include(wagtail_urls))
#new
path(
"",
TemplateView.as_view(template_name="pages/newhomepage.html"),
name="newhomepage",
),
However I would like the blogindex page available at e.g. myste.com/blog but I am not sure how to go about this. Adding the following to urls.py does not do it
path("blog/", include(wagtail_urls))
There are a couple of changes in project required to achieve this.
Firstly, create a urls.py file inside your specific app, this will be different from the one that you should already have alongside wsgi.py, asgi.py, etc. and it will only store app specific urls.
from django.urls import path
from .views import index
urlpatterns = [
path('test/', index, name='generalized-test-url'),
]
Now, you should render the template inside your view.
def index(request):
return render(request, 'pages/newhomepage.html', context)
Lastly, in your root urls.py file, you need to include the app urls. Lets say your app name is blog.
from django.urls import path, include
urlpatterns = [
path('blog/', include('blog.urls')),
]
I have an app which is called sitepages which have three models I wanna pass their objects to a html page called listing.html which is inside templates/sitepages
the sitepages urls.py contains:
from django.views.generic import TemplateView
from django.urls import include
from django.urls import path
from sitepages import views
app_name = "sitepages"
urlpatterns = [
path("news-events-listing/", views.view, name='listing'),
]
the sitepages/views.py:
from .models import Listing, Details1, Details2
from django.template.response import TemplateResponse
def view(request):
return TemplateResponse(request, 'sitepages/listing.html', {
'first_pages': Details1.objects.all(),
'seconde_pages': Details1.objects.all(),
'listing_page': Listing.objects.first(),
})
in the root urls.py I added:
path("", include("sitepages.urls", namespace='sitepages'))
when I put in any template the following:
Listing
it redirects me to the url /news-events-listing but no page is found and it gives me 404 error...what am I doing wrong? why the template is not returned? I should mention that I'm using wagtail for the whole site (I don't know if it's related)
In your root urls.py, make sure your new line
path("", include("sitepages.urls", namespace='sitepages'))
appears before the path("", include(wagtail_urls)) line. The wagtail_urls pattern matches any URL path and passes it to Wagtail to be handled as a Wagtail page, so any patterns after it will never be reached.
I'm very very new to Python 3 and Django and I get to the following problem: I use a standard Template and now how to set it up when there is 1 view. But I don't get the code right for multiple views. I currently run the page locally
At the moment I have tried to change different orders within urlpatterns, and they do work when only 1 url in in there, but I can't get the second one in
views.py
from django.shortcuts import render, render_to_response
# Create your views here.
def index(request):
return render_to_response('index.html')
def store(request):
return render_to_response('store.html')
urls.py
from django.conf.urls import include, url
from django.contrib import admin
from myapp import views as views
from django.contrib.staticfiles.urls import staticfiles_urlpatterns
urlpatterns = [
url(r'^$', views.index, name='index'),
url(r'^store/$', views.store, name='store'),
url(r'^admin/', admin.site.urls)
]
urlpatterns += staticfiles_urlpatterns()
I would like the url pattern that lets me go to the index view and the store view
EDIT:
Full code is shared via: https://github.com/lotwij/DjangoTemplate
The error in the comments shows you are going to http:/127.0.0.1:8000/store.html, but your URL pattern url(r'^store/$', ...) does not include the .html, so you should go to http:/127.0.0.1:8000/store/.
The Django URL system uncouples the URL from the name of the template (sometimes the view doesn't even render a template!). You could change the regex to r'^store.html$ if you really want .html in the URL, but I find the URL without the extension is cleaner.
I have a project with two apps with their respective views, but I want to create a views.py in django project directory for some generic pages such as about, login... It is correct to place the views.py and templates in the root project folder for this purpose?
I have been reading about this. The practice more recommended for this is create an app that creates cohesion between other apps, call it web_site or site whatever is better for you.
python manage.py startapp my_site
Everything in this site app will be done the regular way.
In the projects url you will want to import the url in this way so the web pages appear in the / url pattern.
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('my_site.urls'))
]
I do this for project level pages like you describe. These are usually simple pages that tie the project's apps together, and not contain any business logic.
You can do that by making views.py file into the project directory.
from django.contrib import admin
from django.urls import path, include
from attendance import views
urlpatterns = [
path('admin/', admin.site.urls),
path('', include('insert.urls')),
path('login/', views.logins, name='login'),# relevant line
]
And this is the code of login view
from django.shortcuts import render
from django.http import HttpResponse
# Create your views here.
def logins(request):
return HttpResponse("THIS IS LOGIN PAGE !!!")
In Django, how can I do a simple redirect directly from urls.py? Naturally I am a well organized guy, favoring the DRY principle, so I would like to get the target based on it's named url pattern, rather than hard coding the url.
If you are on Django 1.4 or 1.5, you can do this:
from django.core.urlresolvers import reverse_lazy
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^some-page/$', RedirectView.as_view(url=reverse_lazy('my_named_pattern'), permanent=False)),
...
If you are on Django 1.6 or above, you can do this:
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^some-page/$', RedirectView.as_view(pattern_name='my_named_pattern', permanent=False)),
...
In Django 1.9, the default value of permanent has changed from True to False. Because of this, if you don't specify the permanent keyword argument, you may see this warning:
RemovedInDjango19Warning: Default value of 'RedirectView.permanent' will change from True to False in Django 1.9. Set an explicit value to silence this warning.
This works for me.
from django.views.generic import RedirectView
urlpatterns = patterns('',
url(r'^some-page/$', RedirectView.as_view(url='/')),
...
In above example '/' means it will redirect to index page,
where you can add any url patterns also.
for django v2+
from django.contrib import admin
from django.shortcuts import redirect
from django.urls import path, include
urlpatterns = [
# this example uses named URL 'hola-home' from app named hola
# for more redirect's usage options: https://docs.djangoproject.com/en/2.1/topics/http/shortcuts/
path('', lambda request: redirect('hola/', permanent=False)),
path('hola/', include("hola.urls")),
path('admin/', admin.site.urls),
]
I was trying to redirect all 404s to the home page and the following worked great:
from django.views.generic import RedirectView
# under urlpatterns, added:
url(r'^.*/$', RedirectView.as_view(url='/home/')),
url(r'^$', RedirectView.as_view(url='/home/')),
This way is supported in older versions of django if you cannot support RedirectView
In view.py
def url_redirect(request):
return HttpResponseRedirect("/new_url/")
In the url.py
url(r'^old_url/$', "website.views.url_redirect", name="url-redirect"),
You can make it permanent by using HttpResponsePermanentRedirect
You could do straight on the urls.py just doing something like:
url(r'^story/(?P<pk>\d+)/',
lambda request, pk: HttpResponsePermanentRedirect('/new_story/{pk}'.format(pk=pk)))
Just ensure that you have the new URL ready to receive the redirect!!
Also, pay attention to the kind of redirect, in the example I'm using Permanent Redirect