Django didn't return an HttpResponse object - django

Question

I made a simple pet store app and just added search box feature and I received this error
ValueError at /pet/search/
The view mysite.pet.views.search_page didn't return an HttpResponse object.
I tried to change render_to_response into HttpResponseRedirect but still got the same error.
Linking back to my search_page function in views.
def search_page(request):
form = SearchForm()
if request.method == "POST":
f = SearchForm(request.POST)
if f.is_valid():
Pets = Pet.objects.filter(animal = f.cleaned_data["text"])
return HttpResponseRedirect("search.html",{"Pets":Pets},{"form":form})
else:
return render_to_response("search.html",{"form":form} , context_instance = RequestContext(request))
I did some research and I understand a view has to return a HttpResponse when a HttpRequest is made and render_to_response is just a shortcut.Can someone help explain why this function won't work.Thank you

Answer 1

You are getting this problem because you havn't written a HttpResponse object if request type is not POST
To overcome this in your view write somthing which will process if request type is not post
def search_page(request):
form = SearchForm()
if request.method == "POST":
f = SearchForm(request.POST)
if f.is_valid():
Pets = Pet.objects.filter(animal = f.cleaned_data["text"])
return HttpResponseRedirect("search.html",{"Pets":Pets},{"form":form})
return render_to_response("search.html",{"form":form} , context_instance = RequestContext(request))
Hope this will help you thanks

Answer 2

The error is because when the function is called the method type is not POST and it does not find the corresponding HttpResponse object.
def search_page(request):
form = SearchForm()
if request.method == "POST":
f = SearchForm(request.POST)
if f.is_valid():
Pets = Pet.objects.filter(animal = f.cleaned_data["text"])
return HttpResponseRedirect("search.html",{"Pets":Pets},{"form":form})
else:
return render_to_response("search.html",{"form":form} , context_instance = RequestContext(request))
return render_to_response("any.html",{} , context_instance = RequestContext(request))

Answer 3

def addsponser(request):
if request.method == 'POST':
# return HttpResponse(request,'error is here')
if (request.POST.get('firstname') and
request.POST.get('lastname') and
request.POST.get(' email') and
request.POST.get('phone_Number') and
request.POST.get('gender') and
request.POST.get('state') and
request.POST.get('adress') and
request.POST.get('postal_code') and
request.POST.get('town')
):
fname = request.POST.get('firstname')
lname = request.POST.get('lastname')
em = request.POST.get(' email')
phn = request.POST.get('phone_Number')
gnd = request.POST.get('gender')
stt = request.POST.get('state')
add = request.POST.get('adress')
pstc = request.POST.get('postal_code')
twn = request.POST.get('town')
try:
sponser = Sponsers()
sponser.firstname = fname
sponser.lastname = lname
sponser.email = em
sponser.Phone_Number = phn
sponser.gender = gnd
sponser.state = stt
sponser.adress = add
sponser.postal_code = pstc
sponser.town = twn
sponser.save()
messages.success(request, "sponser Added")
return redirect('sponsers')
except Exception:
messages.error(request, "Failed to add sponser")
return redirect('sponsers')
else:
pass
else:
return redirect('sponsers')