One of the objects I am managing in my django site will only ever have one instance in the database. I therefore want to change the list view to simply redirect to the 'edit' page for this first object.
So basically when you hit /admin/my_site/widgets I want to redirect to /admin/my_site/widget/1. I have tried a custom view, a custom template, etc, but I can't find an easy way of doing this (or any way of doing this for that matter).
It's almost like I want to do something like this (doesn't work because I can't figure out how to change the list view):
class WidgetAdmin(admin.ModelAdmin):
def list_view(self, request):
widget = Widget.objects.all()[0]
return HttpResponseRedirect('/admin/my_site/widget/%s' % widget.id)
I've also tried change the url's to match the list request and do a redirect there, but I can't seem to match the list request with anything other than a complete blanket regex, i.e. (r/^.*$/) which means I just get an infinite loop redirect.
I needed the same thing. I solved it slighty different using the changelist_view from ModelAdmin. Using your example it would look somthing like:
class MySingleEditAdmin(admin.ModelAdmin):
def changelist_view(self, request, extra_context=None):
object, created = self.get_my_object()
url = reverse('admin:%s_%s_change' %(object._meta.app_label, object._meta.module_name), args=[object.id] )
return HttpResponseRedirect(url)
class WidgetAdminAdmin(MySingleEditAdmin):
def get_my_object(self):
return Widget.objects.get_or_create(pk=1, ...default_data...)
Ok this is how I sorted it out.
class WidgetAdmin(admin.ModelAdmin):
def list_view(self, request):
widget = Widget.objects.all()[0]
return HttpResponseRedirect('/admin/my_site/widget/%s' % widget.id)
def get_urls(self):
from django.conf.urls.defaults import *
urls = super(WidgetAdmin, self).get_urls()
my_urls = patterns('',
(r'^$', admin.site.admin_view(self.list_view))
)
return my_urls + urls
Related
I have a url, http://127.0.0.1:8000/lesson/riff-lab/1305/pentab-wow/
When a user navigates to the above url, I want to change it to http://127.0.0.1:8000/lesson/riff-lab/1305/pentab-wow/?d:a3ugm6eyko59qhr/pentab-Track_1.js
The appended part is needed in order to load something that I want to load, but the specifics are not important for this question.
Here's what I have tried.
def my_view(request, pk):
context = {}
page = Page.objects.get(pk=pk)
request.GET._mutable = True
request.GET['?d:%s/%s' % (page.dropbox_key, page.dropbox_js_file_name)] = ""
return render(request, template, context)
Also
def my_view(request, pk):
context = {}
page = Page.objects.get(pk=pk)
request.GET = request.GET.copy()
request.GET['?d:%s/%s' % (page.dropbox_key, page.dropbox_js_file_name)] = ""
return render(request, template, context)
These do not change the url.
Can anyone help? Thanks in advance.
You trying to change the aim of a shell that already has hit it's target.
URL comes first, then view routed to it is processed, there is no way to change it without making another request, e.g. returning a redirect response "please open this url now" to the client.
You can easily find it in django docs by this keywords, but what you are trying to do generally doesn't look very reasonable, if you know beforehand how you need to construct url, why change it mid-way or why change it at all if view has all required data? I don't know your context, but it's probable that you need to reconsider your approach.
This is a view for get all the records in the EducationalRecord model:
def all_education_resume(request):
RESUME_INFO['view'] = 'education'
educations_resume = EducationalRecord.objects.all().order_by('-created_date')
template = 'resumes/all_resume.html'
context = {'educations_resume': educations_resume, 'resume_info': RESUME_INFO}
return render(request, template, context)
Now, if I want to write exactly this view for other models (like job resumes, research resumes , etc.),
I must another view one separately.
My question is:
How can I get a view for all these requests, so first check the URL of
the request and then do the relevant query? How can I control URL
requests in my views?
My other question is exactly the same as my first question,with this difference:
control view that must render in specific template.In other words,in
second question the ratio between the template and the view is instead
of the ratio of the view to the url or how to create a template for
multiple views (for example, for a variety of database resume
resumes, I have a template) and then, depending on which view render,
the template output is different.
I have implemented these two issues as follows:
I wrote a view for each of request!
In each view, I set the value of RESUME_INFO['view'], and then I've checked it in a template page and specified the corresponding template.
What is the best solution to these two questions?
How can I get a view for all these requests, so first check the URL of the request and then do the relevant query? How can I control URL requests in my views?
You can access request.path, or you can let the url(..)s pass a parameter with kwargs that holds a reference to the model for example, but this is usually bad design. Typically if you use different models, you will likely have to order these different as well, filter these differently, render these differently, etc. If not, then this typically indicates that something is wrong with the modeling.
You can however make use of class-based views [Django-doc], to remove as much boilerplate as posssible. Your view looks like a ListView [Django-doc], by using such view, and patching where necessary, we can omit most of the "boilerplate" code:
# app/views.py
from django.views.generic.list import ListView
class MyBaseListView(ListView):
resume_info = None
template = 'resumes/all_resume.html'
def get_context_data(self, *args, **kwargs):
context = super().get_context_data(*args, **kwargs)
context['resume_info'] = {'view': self.resume_info}
return context
In the individual listviews, you then only need to specify the resume_info and the model or queryset to render it with the 'all_resume.html' template, for example:
# app/views.py
# ...
class EducationalResumeView(MyBaseListView):
queryset = EducationalRecord.objects.order_by('-created_date')
resume_info = 'education'
class OtherModelView(MyBaseListView):
model = OtherModel
resume_info = 'other_info'
So we can here use inheritance to define common things only once, and use it in multiple views. In case we need to change something in a specific view, we can override it at that level.
In the urls.py, you define such view with the .as_view() method [Django-doc]. For example:
# app/urls.py
from django.urls import path
from app.views import EducationalResumeView, OtherModelView
urlpatterns = [
path('education/', EducationalResumeView.as_view()),
path('other/', OtherModelView.as_view()),
]
According to django docs:
class MyModelAdmin(admin.ModelAdmin):
def get_urls(self):
urls = super(MyModelAdmin, self).get_urls()
my_urls = [
url(r'^my_view/$', self.my_view),
]
return my_urls + urls
def my_view(self, request):
# ...
context = dict(
# Include common variables for rendering the admin template.
self.admin_site.each_context(request),
# Anything else you want in the context...
key=value,
)
return TemplateResponse(request, "sometemplate.html", context)
If I am not wrong, we can do the same thing by adding url in urls.py and the views in views.py as it is normally done then, what is the use of introducing this way? I am a newbie to django and I may be missing something here.
Can you please provide an example where we cannot do it in views.py and we must use the above method?
Any guidance/help would be appreciated.
I think I figured out, both of them can be used to do the same thing but the key difference is that the views which you write using above method will belong to admin app while the general views in views.py belongs to the particular app in you have written.
Hence, the url in ModelAdmin need to be called using name admin:url_name since the url goes as admin/my_views/ in given example.
Have the next Django REST question.
I have the view.
class MessageViewSet(viewsets.ModelViewSet):
serializer_class = MessageSerializer
queryset = Message.objects.filter(isread = False)
def mark_read():
queryset = Message.objects.update(isread=True)
return Response({'read':queryset})
And router in urls.py
router = SimpleRouter() router.register(r'api/get_messages', MessageViewSet)
urlpatterns = [
url(r'^$', MainView.as_view(), name='main'),
url(r'^', include(router.urls)) ]
Now i have 'get_messages' page which shows all list.
How can i implement a method which would change 'isread' value of model instanse from False to True, when I visit a 'mark_read' page?
As you can see, i tried to write method in the class. But when i'm trying to call it in urls in this way:
router.register(r'api/mark_read', MessageViewSet.mark_read),
Here comes an error.
assert queryset is not None, 'base_name argument not specified, and could ' \
AssertionError: base_name argument not specified, and could not automatically determine the name from the viewset, as it does not have a .queryset attribute.
Maybe i shouldnt use router, and rewrite view and urls in other way. If u know how to solve this problem, please answer. Thanks.
You can use detail_route or list_route decorators.
from rest_framework.decorators import list_route
class MessageViewSet(viewsets.ModelViewSet):
#list_route()
def mark_read(self, request):
queryset = Message.objects.update(isread=True)
return Response({'read':queryset})
With that mark_read method will be available at api/get_messages/mark_read. And you don't need to create separate router, just use one you created for MessageViewSet
docs reference
Since you are using a model viewset you can directly use put or patch rest method to send the desired value for the desired field as the data.
Ideally in rest get should not change model values. If you really want a different end point put the list_route or detail_route decorator on your mark_read method, and make them a valid call for only a put and/or patch call
from rest_framework.decorators import list_route
class MessageViewSet(viewsets.ModelViewSet):
#list_route(methods=['Patch', 'PUT'])
def mark_read(self, request):
queryset = Message.objects.update(isread=True)
return Response({'read':queryset})
Thanks to #ivan-semochkin and #Shaumux for replies. Advices were really helpful.
That is my route. I used detail_route instead of list_route.
#detail_route(methods=['get','put'], url_name='mark_read/')
def mark_read(self, request, pk=None):
queryset = Message.objects.filter(pk=pk).update(isread=True)
return Response({'read':queryset})
Now 'isread' value is changing wnen i visit 'mark_read' page.
Link: "api/get_messages/pk/mark_read"
Does anyone know, is it posslible to make links looking the next way:
"api/get_messages" - list, "api/mark_read/pk" - changing isread value.
Is it possible to create something like this? "api/mark_read?=pk"
I'm trying to add a custom page to the admin without a model association.
This is what I achieved so far.
class MyCustomAdmin(AdminSite):
def get_urls(self):
from django.conf.urls import url
urls = super(MyCustomAdmin, self).get_urls()
urls += [
url(r'^my_custom_view/$', self.admin_view(MyCustomView.as_view()))
]
return urls
class MyCustomView(View):
template_name = 'admin/myapp/views/my_custom_template.html'
def get(self, request):
return render(request, self.template_name, {})
def post(self, request):
# Do something
pass
admin_site = MyCustomAdmin()
admin_site.register(MyModel1)
admin_site.register(MyModel2)
# etc...
This is actually working but the problem is that with this solution I loose some apps from the Django admin interface (account, auth, socialaccounts, sites).
This is because your other admins are using the default admin.site. You need to totally replace the default admin.site with your own as explained here (you may also want to read this too).
Or you can just do it piggy-style by monkeypatching the default admin.site.get_urls() method:
from django.contrib import admin
_admin_site_get_urls = admin.site.get_urls
def get_urls():
from django.conf.urls import url
urls = _admin_site_get_urls()
urls += [
url(r'^my_custom_view/$',
admin.site.admin_view(MyCustomView.as_view()))
]
return urls
admin.site.get_urls = get_urls
Legal disclaimer : I won't be held responsible for any kind of any unwanted side-effect of this "solution", including (but not restricted too) your coworkers defenestrating you on the next code review. It's a dirty solution. It's a mess. It stinks. It's evil. You shouldn't do that, really.