Djangoで1人のビュー生成を許可する

Question

this question以下のコードを思いついた。

次のように動作は次のようになります。

1. 各Suggestionが

許可されているのみ1 Voteユーザーあたりのそれは正常に動作しますが、私は最善の方法だと思いまし

Voteいくつかを受け取ることができますこれをする？

vote_form.html

<form action="{% url 'add_vote' suggestion.id %}" method="post"> 
    {% csrf_token %} 
    <input type="submit" value="I want to vote"> 
</form> 

urls.py

urlpatterns = [ 
    url(r'^suggestion/(?P<pk>\d+)/$', views.SuggestionDetail.as_view(), name="suggestion_detail"), 
    url(r'^suggestion/(?P<pk>\d+)/vote/$', views.add_vote, name='add_vote'), 
] 

models.py

class Vote(models.Model): 
    suggestion = models.ForeignKey(Suggestion) 
    voter = models.ForeignKey('auth.User') 
    vote_count = models.BooleanField() 

views.py

def add_vote(request, pk): 
    suggestion = get_object_or_404(Suggestion, pk=pk) 
    vote = Vote(
     suggestion = suggestion, 
     voter = request.user, 
     vote_count = True) 
    has_user_voted = Vote.objects.filter(voter=request.user, suggestion=suggestion).count() 
    if has_user_voted < 1: 
     vote.save() 
    else: 
     messages.error(request, 'It seems you have already voted, only one vote is allowed') 
    return HttpResponseRedirect(reverse('suggestion_detail', args=(suggestion.id,))) 

Answer 1

最初のソリューションは、あなたのデータにTrueあなたもget-or-create

def add_vote(request, pk): 
    suggestion = get_object_or_404(Suggestion, pk=pk) 
    vote, created = Vote.objects.get_or_create(
      voter=request.user, 
      suggestion=suggestion, 
      vote_count=True) 
    if not created: 
     messages.error(request, 'It seems you have already voted, only one vote is allowed') 
    return HttpResponseRedirect(reverse('suggestion_detail', args=(suggestion.id,)))