django-views

Just received the Sentry error TypeError context must be a dict rather than Context. on one of my forms. I know it has something to do with Django 1.11, but I am not sure what to change to fix it. Offending line message = get_template('email_forms/direct_donation_form_email.html').render(Context(ctx)) Entire View def donation_application(request): if request.method == 'POST': form = DirectDonationForm(data=request.POST) if form.is_valid(): stripe.api_key = settings.STRIPE_SECRET_KEY name = request.POST.get('name', '') address = request.POST.get('address', '') city = request.POST.get('city', ''

How do I use more than one form per page in Django? Herman Schaaf Please see the following previously asked (and answered) questions: Django: multiple models in one template using forms and Proper way to handle multiple forms on one page in Django . Depending on what you're really asking, this is my favorite way to handle different models on the same page: if request.POST(): a_valid = formA.is_valid() b_valid = formB.is_valid() c_valid = formC.is_valid() # we do this since 'and' short circuits and we want to check to whole page for form errors if a_valid and b_valid and c_valid: a = formA.save

I'm having a problem with logged users and a Django ModelForm . I have a class named _Animal_ that has a ForeignKey to User and some data related to the animal like age, race, and so on. A user can add Animals to the db and I have to track the author of each animal, so I need to add the request.user that is logged when the user creates an animal instance. models.py class Animal(models.Model): name = models.CharField(max_length=300) age = models.PositiveSmallIntegerField() race = models.ForeignKey(Race) ... publisher = models.ForeignKey(User) def __unicode__(self): return self.name class

Is there a way to get request.session from inside a class-based view? For instance, I have from django.views.generic.edit import FormView class CreateProfileView(FormView): def form_valid(self, form): # --> would like to save form contents to session here return redirect(self.get_success_url()) The only thing I can think of would be override as_view by adding def as_view(self, request, *args, **kwargs): self.session = request.session super(CreateProfileView, self).as_view(request, *args, **kwargs) to the class. But that seems ugly. Is there another way? Timmy O'Mahony You have access to self

EMP_CHOICES = ( (0,'-- Select --'), (1,'Good'), (2,'Average'), ) class EMPFeedback(models.Model): user_choices = models.IntegerField(choices=EMP_CHOICES) If the value stored in the db as 1 for user_choices how to print the corresponding user_choices corresponding value (i.e. 1==GOOD) fb = EMPFeedback.objects.get(id=1) print fb.user_choices # prints 1 print fb.user_choices.EMP_CHOICES There's a method for that! (™ Apple) fb.get_user_choices_display() 来源： https://stackoverflow.com/questions/4274243/django-print-choices-value

hello i'm new in python and django I need a view that get current user profile I know I shoud use get_profile from User but I don't know how to use it . i read the django document and It didn't help me. this is what I found from doc: from django.contrib.auth.models import User profile=request.user.get_profile() César Django's documentation says it all, specifically the part Storing additional information about users . First you need to define a model somewhere in your models.py with fields for the additional information of the user: models.py from django.contrib.auth.models import User class

Is possible to add GET variables in a redirect ? (Without having to modifiy my urls.py) If I do redirect('url-name', x) I get HttpResponseRedirect('/my_long_url/%s/', x) I don't have complains using HttpResponseRedirect('/my_long_url/%s/?q=something', x) instead, but just wondering... Is possible to add GET variables in a redirect ? (Without having to modifiy my urls.py) I don't know of any way to do this without modifying the urls.py . I don't have complains using HttpResponseRedirect('/my_long_url/%s/?q=something', x) instead, but just wondering... You might want to write a thin wrapper to

This is related to this question: Django return json and html depending on client python I have a command line Python API for a Django app. When I access the app through the API it should return JSON and with a browser it should return HTML. I can use different URLs to access the different versions but how do I render the HTML template and JSON in the views.py with just one template? To render the HTML I would use: return render_to_response('sample/sample.html....') But how would I do the same for JSON without putting a JSON template? (the content-type should be application/json instead of