i wanna redirect to success page but it seems i should use different way in post request response.
views.py :
def uploader(request):
if request.method == 'POST':
form = UploadFileForm(request.POST, request.FILES)
if form.is_valid():
return redirect("download",link = 'test')
else:
form = UploadFileForm()
return render(request, 'uploader/pages/upload.html', {'form': form})
def downloader(request,link) :
return render(request, 'uploader/pages/download.html',)
urls.py :
urlpatterns = [
re_path(r'^admin/', admin.site.urls),
re_path(r'^$', views.uploader, name='upload'),
re_path(r'^(?P<link>[\w-]+)/', views.downloader, name='download'),
]
log in terminal :
[27/Jul/2024 17:43:37] "POST / HTTP/1.1" 302 0
[27/Jul/2024 17:43:37] "GET /test/ HTTP/1.1" 200 46
I've tried HttpResponseRedirect('test')
and HttpResponseRedirect(reverse('download', kwargs={'link': 'link'}))
to
After correct questions in comments I understand.
You use fetch
on client-side to made request to back-end per javascript. It is not Django resposibility, but I try to answer.
After fetch request on client-side, you should initiate a programmatic redirect on client-side if you receive a redirect response.
window.location.repelace(response.headers.get('location'))
But if we read documentation about fetch - we can see, we can do redirect automatically, add in fetch additional attribute:
fetch (url, { method: 'POST', redirect: 'follow'})
more here: https://developer.mozilla.org/en-US/docs/Web/API/RequestInit#redirect