Django low level cache views

2018-06-02 16:34:40

I have an index view which validates a form containing various data. Even though the thankyou.html page doesn't have complex calculations to kill the server, i would like to render aa slighlty different html page if thankyou.html is already low level cached. To tell you the truth, I don't know what key to pass it... Here is the code.

def index(request):

form = UserForm()

message = 'Incorrect data!'

if request.method == 'POST':

form = UserForm(request.POST)

if form.is_valid():

try:

User.objects.get(code=form.cleaned_data['code'])

except (KeyError, ObjectDoesNotExist):

u = User(lastname=form.cleaned_data['lastname'], surname=form.cleaned_data['surname'], address=form.cleaned_data['address'], email=form.cleaned_data['email'], phone=form.cleaned_data['phone'], code=form.cleaned_data['code'], )

u.save()

return HttpResponseRedirect('/thanks/')

#return redirect('thankyou')

return render_to_response('index.html',{'message': message,'form' : form}, context_instance=RequestContext(request)).

I guess this is the way I should low level cache it:

if form.is_valid():

key = ???

cached_html = cache.get (key)

try:

User.objects.get(code=form.cleaned_data['code'])

except (KeyError, ObjectDoesNotExist):

if not cached_html:

cached_html = render_to_response('ty.html',{ }, context_instance=RequestContext(request))

cache.set(key, cached_html, time_until_midnight())

return HttpResponseRedirect('/thanks/')

#return redirect('thankyou')

我认为在这种情况下，你应该使用字符串（'thankyou'+ form.cleaned_data ['code']）作为键

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