Question

I have class:

class Image(models.Model):
    title = models.CharField(max_length=100)
    imgFile1 = models.FileField(upload_to='%Y/%m/%d')
    vote1 = models.IntegerField(default=0)
    trofeum1 = models.IntegerField(default=0)
    imgFile2 = models.FileField(upload_to='%Y/%m/%d')
    vote2 = models.IntegerField(default=0)
    trofeum2 = models.IntegerField(default=0)
    home = models.IntegerField(default=0)

and function in views.py:

def vote1(request,i_id=0):
    imgId = str(i_id)
    try:
        if (request.session[imgId] == True):
            return HttpResponseRedirect(reverse('app.views.err'))              
    except Exception:
        request.session[imgId] = True
    try:
        image = Image.objects.get(pk=i_id)
    except Exception:
        image = None
    image.vote1+=1
    image.save()
    if image.vote1 == 5:
        image.trofeum1 += 1
        image.trofeum2 = 0
        image.vote1 = 0
        image.vote2 = 0
        image.imgFile2 = None
        image.save()
    return HttpResponseRedirect('app.views.index'))

So it updates certain object in my database after clicking some link on my site, but I dont want to redirect after that, just want to change object in my database – CLICK->UPDATE->NOTHING ELSE. Do you know any solutions to avoid error:The view app.views.vote1 didn’t return an HttpResponse object or any other way to do that?

I redirect to the same page but after that I’m on the top of the page – I want to be on the same place where I clicked the link.

Answer 1

Try to do the same thing with Ajax. It will do that.