I have an app that shows a list of "Issues". The main urls.py file sends /issues/ to the urls.py file in the "issues" app.
urlpatterns = [
path('', RedirectView.as_view(url='/issues/')),
path('admin/', admin.site.urls),
path('issues/', include('issues.urls')),
]
In the issues app's urls.py file I have:
path('', views.IssueListView.as_view(), name='issue-list'),
That calls the IssueListView which is a generic ListView view:
class IssueListView(generic.ListView):
model = Issue
Now, I want to add a sidebar menu with links users can click to sort the issues list by category. I understand that I can rewrite the get_queryset() method of the IssueListView to accept a kwarg, and just load all the issues if that kwarg is missing (with an if statement that checks for the presence of the kwarg, right?), but I think to do this I need to have two urlpatterns that point to the same view like:
path('', views.IssueListView.as_view(), name='issue-list'),
path('<category>', views.IssueListView.as_view(), name='issue-category-list'),
But I am wondering if that's the normal "Django way" to do it.
Thank you.
In the case you have specified, you can use the same view for both url pattern. There is nothing wrong in that.
path('', views.IssueListView.as_view(), name='issue-list'),
path('<category>', views.IssueListView.as_view(), name='issue-category-list'),
Related
Here is the problem:
I have an app with the following models: project, position, outreach
A position is connected to a project and project only with a Foreign key
An outreach is connected to a position and a position only with a Foreign key
I can create a new project from almost anywhere in my app (same for the other objects). Currently I wrote that a new project is created from the url dashboard/newjobproject but I would to make it so that depending on the page that I am, the url simply becomes something like www.myapp.com/..../newproject
What's a way to write the urls.py to achieve that?
from django.urls import path
from action import views
app_name = 'action'
urlpatterns = [
# ex: /action/
path('', views.login, name='login'),
path('dashboard/', views.dashboard, name='dashboard'),
path('contacts/', views.contacts, name='contacts'),
path('projects/', views.project, name='project'),
path('contacts/newcontact', views.new_contact, name='new_contact'),
path('projects/newjobproject', views.new_outreach, name='new_outreach'),
path('dashboard/newjobproject', views.new_jobproject, name='new_jobproject'),
path('projects/<uuid>/newjobposition', views.new_jobposition, name='new_jobposition'),
]
However,
Try adding this to the bottom of urlpatterns:
path('<path:p>/newjobproject', views.new_jobproject, name='whatever-name-you-want'),
and in your views.py:
def new_jobproject(request, p):
Tbh though, this is sort of a hacky way to do it. It'll break in a few locations. If you have a main urlpatterns array in which you're including the urls for this 'action' app as APIs, this solution won't work outside the API urls.
For eg. if you have:
urlpatterns = [
path('admin/', admin.site.urls),
path('api/v1/', include('action.urls')),
]
And you access your url like this -> www.myapp.com/api/v1/....../newjobproject, then the code will work. If you try to access www.myapp.com/..../newjobproject or www.myapp.com/admin/..../newjobproject then the code will break and it will not match any of the paths. I'm not sure how you'd get it to work in that case.
If the above scenario is not an issue, if you're not going to be using these views as APIs, and your urlpatterns looks something like this:
urlpatterns = [
path('admin', admin.site.urls),
path('/', include('action.urls')),
]
then the code should work for all urls except for the admin/.../newjobproject case.
I am learning Django and am having an issue with some simple testing and passing URL parameters. Here is my urls.py:
urlpatterns = [
path('admin/', admin.site.urls),
path('', views.home, name='home'),
path('custnames', views.custnames, name='custnames'),
path('custdetail/<int:cust_id>/', views.cust_detail, name='cust_detail'),
]
And here is my views.py
def cust_detail(request, cust_id):
return HttpResponse('<p>Cust Detail View with cust_id {cust_id}</p>')
When I put this for my URL in my browser: http://localhost:8000/custdetail/1/
My output is:
Cust Detail View with cust_id {cust_id}
The "home" and "custnames" sections seem to work fine.
Any ideas on what I'm doing wrong here?
~Ed
Thanks to "minglyu" and "Melvyn". Adding the "f" worked.
This is how I defined urls.py file of my app
router = DefaultRouter()
router.register('hello-viewset', views.HelloViewSet, base_name='hello-viewset')
router.register('profiles', views.UserProfileViewSet)
router.register('schema', views.SchemaViewSet)
router.register('creddefination', views.CredDefViewSet)
router.register('overalltable', views.OverallViewSet)
urlpatterns = [
path('', include(router.urls)),
]
urls.py of Project:
urlpatterns = [
path('admin/', admin.site.urls),
path('api/', include('DIAPI.urls')),
]
I am not getting correct address for creddefination. But when i manually go to http://127.0.0.1:7000/api/creddefination/ it is working. It is just not displaying correctly. What might be reason for this
I guess views.CredDefViewSet and views.OverallViewSet are using the same model.
If that's true, then the default register's basename will be named after that model and used as name in a call to Django's reverse url construction. Since the API Root view will be trying to resolve both views with the same name, it'll lead to the same url.
Workaround is to explicitly add a basename to one of the view:
router.register('creddefination', views.CredDefViewSet, basename='creddeef')
Was hoping someone could point me in the right direction with this. I've tried nearly everything I can think of, but I can't seem to get it to work. I've got a set of URLs I'd like to match in Django:
www.something.com/django/tabs/
www.something.com/django/tabs/?id=1
Basically, I want to make it so that when you just visit www.something.com/django/tabs/ it takes you to a splash page where you can browse through stuff. When you visit the second URL however, it takes you to a specific page which you can browse to from the first URL. This page is rendered based on an object in the database, which is why the id number is there. I've tried to account for this in the URL regex, but nothing I try seems to work. They all just take me to the main page.
This is what I have in urls.py within the main site folder:
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^tabs/', include("tabs.urls")),
]
and within urls.py in the app's folder:
urlpatterns = [
url(r'\?id=\d+$', tab),
url(r'^$', alltabs)
]
Would anyone be so kind as to point me in the right direction? Thanks in advance!
You are not following the right approach here. Query paramers are used to change the behaviour of the page slightly. Like a added filter, search query etc.
What i would suggest is you have only one view and render different templates based on query parameters in the view.
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^tabs/', alltabs),
]
In your alltab views you can have something like this.
def alltabs(request):
if request.GET.get("id"):
id = request.GET.get("id")
your_object = MyModel.objects.get(id=id)
return render_to_response("tab.html", {"object":your_object})
return render_to_response("alltab.html")
Hope this helps
This is not the preferred 'django way' of defining urls patterns, I would say:-)
In the spirit of django would be something like
www.something.com/django/tabs/
www.something.com/django/tabs/1/
....
www.something.com/django/tabs/4/
and for this you define your url patterns within the app for example this way
tabs/urls.py:
from django.conf.urls import url
from . import views
urlpatterns = [
# ex: /tabs/
url(r'^$', views.index, name='index'),
# ex: /tabs/5/
url(r'^(?P<tab_id>[0-9]+)/$', views.detail, name='detail'),
# ex: /tabs/5/results/
url(r'^(?P<tab_id>[0-9]+)/results/$', views.results, name='results'),
]
and something similar in your views
tabs/views.py:
from django.shortcuts import get_object_or_404, render
from tabs.models import Tab
def index(request):
return render(request, 'tabs/index.html')
def detail(request, tab_id):
tab = get_object_or_404(Tab, pk=tab_id)
return render(request, 'tabs/detail.html', {'tab': tab})
...
You can follow this django tutorial for more details:
My code is as follows:
root urls.py
from django.conf.urls import include, url
from django.contrib import admin
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'', include('app.urls')),
]
My applications url is like so
urlpatterns = [
url(r'^$',views.login, name='login'),
url(r'^homepage/$', views.homepage, name='homepage'),
]
The first screen the user will see is the login screen (views.login). At the moment I just want to set the login button to be a url that takes them to the homepage (just for practice) but it doesnt seem to work.
The login html is like so
<button type="button">Log-In</button>
This should go to my urls page above...find the name 'homepage' and take me to views.homepage which is as so:
def homepage(request):
return render(request, 'application/homepage.html', {})
but my homepage doesnt get rendered and I have absolutely no idea why its driving me crazy.
Any help would be appreciated.
This is nothing to do with Django, but a pure HTML problem. You can't put a link inside a button. A button needs to be part of a form, and submits to the action value of that form. Either do that, and take the a tag out; or, remove the button, and just use the a.